5ach14(arithmetic And Geometric Sequences And Their Summation)

Certificate Mathematics in Action Full Solutions 5A

14 Arithmetic and Geometric Sequences and their Summation • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • (c) very large

Activity Activity 14.1 (p. 161) 1.

Follow-up Exercise

(a) 2, 2, 2, 2

p. 158

(b) 3, 3, 3, 3

1.

(c) 5, 5, 5, 5 (d) –5, –5, –5, –5 2.

T(3) = 5(3) = 15 T(4) = 5(4) = 20

They are equal.

Activity 14.2 (p. 184) 1.

(b) T(1) = 12 – 1 = 0 T(2) = 22 – 1 = 3

(a) S(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

T(3) = 32 – 1 = 8 T(4) = 42 – 1= 15

(b) S(10) = 55 2.

No

1

1 1 (c) T(1) =   = 3 3

Activity 14.3 (p. 198) 1 .

Range of R

R

n 8

R < –1

–2 (or any other reasonable answers)

15 very large

–1 < R < 1

1 2 (or any other reasonable answers)

8

15 very large

R>1

2.

(a) T(1) = 5(1) = 5 T(2) = 5(2) = 10

2 (or any other reasonable answers)

(a) very large or very small

8 15 very large

Rn 256 (or any other reasonable answers) –32 768 (or any other reasonable answers) very large or very small 1 (or any 256 other reasonable answers) 1 (or any 32 768 other reasonable answers) very close to zero 256 (or any other reasonable answers) 32 768 (or any other reasonable answers) very large

2

1 1 T(2) =   = 9 3 3

1 1 T(3) =   = 27 3 4

1 1 T(4) =   = 3 81   2.

(a) (i) 21, 26 (ii) ∵ T(1) = 1 = 5(1) – 4 T(2) = 6 = 5(2) – 4 T(3) = 11 = 5(3) – 4 T(4) = 16 = 5(4) – 4 ∴ T(n) = 5n − 4 (iii) ∵ ∴ and ∴

T(n) = 5n – 4 T(10) = 5(10) – 4 = 46 T(15) = 5(15) – 4 = 71 The 10th term and the 15th term of the sequence are 46 and 71 respectively.

(b) (i) 81, 243 (ii) ∵ T(1) = 1 = 31 – 1 T(2) = 3 = 32 – 1 T(3) = 9 = 33 – 1 T(4) = 27 = 34 – 1 n −1 ∴ T(n) = 3

(b) very close to zero (iii) ∵ T(n) = 3n – 1

Certificate Mathematics in Action Full Solutions 5A

∴ T(10) = 310 – 1 = 39 and T(15) = 315 – 1 = 314 ∴ The 10th term and the 15th term of the sequence are 39 and 314 respectively. (c) (i)

1 1 , 32 64

= 4 log 2 – 3 log 2 = log 2 ∴ It is an arithmetic sequence with common difference log 2. 2.

1 1 = 2 21 1 1 T(2) = = 2 4 2 1 1 T(3) = = 3 8 2 1 1 = T(4) = 16 2 4 1 ∴ T(n) = n 2

(ii) ∵ T(1) =

(b) Let a and d be the first term and the common difference respectively. ∵ a = 10 and d = 14 – 10 = 4 ∴ T(n) = a + (n – 1)d = 10 + (n – 1)(4) = 6 + 4n ∴ T(12) = 6 + 4(12) = 54

1 2n 1 1 ∴ T(10) = 10 = 2 1024 1 1 and T(15) = 15 = 2 32 768 ∴ The 10th term and the 15th term of the 1 1 sequence are and 1024 32 768 respectively.

(iii) ∵ T(n) =

(c) Let a and d be the first term and the common difference respectively. ∵ a = 6 and d = 4 – 6 = –2 ∴ T(n) = a + (n – 1)d = 6 + (n – 1)(–2) = 8 – 2n ∴ T(12) = 8 – 2(12) = −16 (d) Let a and d be the first term and the common difference respectively. ∵ a = –25 and d = –22 – (–25) = 3 ∴ T(n) = a + (n – 1)d = –25 + (n – 1)(3) = 3n – 28 ∴ T(12) = 3(12) – 28 =8

p. 165 1.

(a) T(2) – T(1) = 3 – 1 = 2 T(3) – T(2) = 5 – 3 = 2 T(4) – T(3) = 7 – 5 = 2 ∴ It is an arithmetic sequence with common difference 2. (b) T(2) – T(1) = –10 – (–13) = 3 T(3) – T(2) = –7 – (–10) = 3 T(4) – T(3) = –4 – (–7) = 3 ∴ It is an arithmetic sequence with common difference 3. (c) T(2) – T(1) = 4 – 2 = 2 T(3) – T(2) = 8 – 4 = 4 ≠ 2 ∴ It is not an arithmetic sequence. (d)

T(2) – T(1) = log 4 – log 2 = 2 log 2 – log 2 = log 2 T(3) – T(2) = log 8 – log 4 = 3 log 2 – 2 log 2 = log 2 T(4) – T(3) = log 16 – log 8

(a) Let a and d be the first term and the common difference respectively. ∵ a = 2 and d = 5 – 2 = 3 ∴ T(n) = a + (n – 1)d = 2 + (n – 1)(3) = 3n – 1 ∴ T(12) = 3(12) – 1 = 35

3.

(a) Let a and d be the first term and the common difference respectively. T(9) = a + 8d = 22 ……(1) T(13) = a + 12d = 34 ……(2) (2) – (1), 4d = 12 d=3 By substituting d = 3 into (1), we have a + 8(3) = 22 a = –2 ∴ The first term and the common difference are –2 and 3 respectively. (b) T(n) = a + (n – 1)d = –2 + (n – 1)(3) = −5 + 3n (c) ∵ T(k) = 73

14

Arithmetic and Geometric Sequences and their Summation

∴ –5 + 3k = 73 k = 26 4.

(a) T(n) = a + (n – 1)d = –24 + (n – 1)(5) = −29 + 5n

∴ The five required arithmetic means are 7, 9, 11, 13 and 15. 3.

(b) T(7) = – 29 + 5(7) =6 T(12) = – 29 + 5(12) = 31 (c) ∵ The mth term is the first positive term of the sequence. ∴ T(m) > 0 i.e. – 29 + 5m > 0 29 m> 5 ∵ m is the number of terms, it must be an integer. ∴ m =6

∵ x is the arithmetic mean between 6 and y, and 18 is the arithmetic mean between y and 22. 6 + y .........(1)   x= 2 ∴ y + 22 18 = .........(2) 2  From (2), we have 36 = y + 22 y = 14 By substituting y = 14 into (1), we have 6 + 14 x= 2 = 10

p. 176 1.

(a)

p. 170 1.

− 10 + (−2) 2 = –6

(a) Arithmetic mean =

(b)

137 + 27 2 = 82

(b) Arithmetic mean =

2.

(a) Let d1 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 5, 5 + d1, 5 + 2d1, 17 ∵ The 4th term is also given by 5 + 3d1. ∴ 5 + 3d1 = 17 d1 = 4 ∴ The two required arithmetic means are 9 and 13.

(c)

(b) Let d2 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 5, 5 + d2, 5 + 2d2, 5 + 3d2, 17 ∵ The 5th term is also given by 5 + 4d2. ∴ 5 + 4d2 = 17 d2 = 3 ∴ The three required arithmetic means are 8, 11 and 14. (c) Let d3 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 5, 5 + d3, 5 + 2d3, 5 + 3d3, 5 + 4d3, 5 + 5d3, 17 ∵ The 7th term is also given by 5 + 6d2. ∴ 5 + 6d3 = 17 d3 = 2 80

(d)

2.

T (2) 4 = =2 T (1) 2 T (3) 6 3 = = ≠2 T (2) 4 2 ∴ It is not a geometric sequence. T (2) −4 1 = = T (1) − 8 2 T (3) −2 1 = = T (2) − 4 2 T (4) −1 1 = = T (3) − 2 2 ∴ It is a geometric sequence with common ratio 1 . 2 T (2) 0.22 11 = = T (1) 0.2 10 T (3) 0.222 111 11 = = ≠ T (2) 0.22 110 10 ∴ It is not a geometric sequence. T (2) log 9 = =2 T (1) log 3 T (3) log 27 3 = = ≠2 T (2) log 9 2 ∴ It is not a geometric sequence.

(a) Let a and R be the first term and the common ratio respectively. 2 ∵ a = 1 and R = = 2 1 ∴ T(n) = aRn – 1 = 1(2)n – 1 = 2n – 1

Certificate Mathematics in Action Full Solutions 5A

∴ T(6) = 26 – 1 = 32 (b)

(c)

(d)

4.

Let a and R be the first term and the common ratio respectively. 9 ∵ a = 3 and R = = 3 3 ∴ T(n) = aRn – 1 = 3(3)n – 1 = 3n ∴ T(6) = 36 = 729 Let a and R be the first term and the common ratio respectively. 6 ∵ a = –3 and R = = –2 −3 n–1 ∴ T(n) = aR = –3(–2)n – 1 3 = (–2)n 2 3 ∴ T(6) = (–2)6 2 = 96 Let a and R be the first term and the common ratio respectively. −64 1 =− ∵ a = 128 and R = 128 2 ∴ T(n) = aRn – 1  1 = 128  −   2

n −1

 1 = –256  −   2  1 ∴ T(6) = –256  −   2 − 4 = 3.

n

6

(a) Let a and R be the first term and the common ratio respectively. 1 ∵ a = –8, R = and T(n) = aRn – 1 2 1 ∴ T(n) = − 8  2 = − 2 4− n (b) T(4) = –24 – 4 = −1 T(6) = –24 – 6 1 =− 4

n −1

(a) Let a and R be the first term and the common ratio respectively. T(3) = aR2 = 1 ………(1) T(8) = aR7 = 243 ………(2) (2) ÷ (1), R5 = 243 R=3 By substituting R = 3 into (1), we have a(3)2 = 1 1 a= 9 ∴ The first term and the common ratio are

1 and 3 9

respectively. (b) T(n) = aRn – 1 1 = ⋅ 3n – 1 9 =3

n −3

(c) T(7) + T(9) = 37 – 3 + 39 – 3 = 810

p. 181 1.

(a) Geometric mean = 5 × 45 = 15 (b) Geometric mean = − 12 × (−147) = 42

2.

(a) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 1 1 1 2 27 , R, R , 2 2 2 16 1 ∵ The 4th term is also given by R3. 2 1 3 27 R = ∴ 2 16 3 R= 2 ∴ The two required geometric means are

3 9 and . 4 8

(b) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 3 3 3 2 3 3 2 , r, r , r , 8 8 8 8 27 3 ∵ The 5th term is also given by r4. 8 3 4 2 r = ∴ 8 27

14

r=±

∴ The three required geometric means are and 3.

3.

2 3 1 1 , 4 6

1 1 1 1 or − , and − . 9 9 6 4

∵ x, x + 3, x + 9 are in geometric sequence. ∴ x + 3 is the geometric mean between x and x + 9. ∴ (x + 3)2 = x(x + 9) 2 x + 6x + 9 = x2 + 9x 3x = 9 x =3

(a) 1 + 3 + 5 + 7 + 9 (b) 25

2.

(a) 2 + 4 + 8 + 16 + 32 + 64 (b) 126

3.

(a) 1 + 4 + 9 + 16 + 25 + 36 + 49

4.

(b) 140

p. 188 1.

(a) ∵ a = 1, d = 5 – 1 = 4 and n = 10 10 ∴ S(10) = [2(1) + (10 – 1)(4)] 2 = 190

2.

(a) ∵ a = –5, l = 9 and n = 8 8( −5 + 9) ∴ S(8) = 2 16 =

(b) ∵ a = 100, d = –4 and n = 12 12 ∴ S(12) = [2(100) + (12 – 1)( –4)] 2 = 936

80

(a) ∵ a = 1, l = 100 and n = 100 100(1 + 100) ∴ 1 + 2 + … + 100 = 2 = 5050 (b) Let n be the number of terms of the given series. ∵ a = 3, d = 6 – 3 = 3 and l = T(n) = 99 and T(n) = a + (n – 1)d ∴ 99 = 3 + (n – 1)(3) n = 33 33(3 + 99) ∴ The required sum = 2 = 1683

(b) ∵ a = 3, d = 8 – 3 = 5 and n = 12 12 ∴ S(12) = [2(3) + (12 – 1)(5)] 2 = 366 (c) ∵ a = 28, d = 26 – 28 = –2 and n = 15 15 ∴ S(15) = [2(28) + (15 – 1)( –2)] 2 = 210

(a) Let n be the number of terms of the given series. ∵ a = –7, d = –2 – (–7) = 5 and l = T(n) = 103 and T(n) = a + (n – 1)d ∴ 103 = –7 + (n – 1)(5) n – 1 = 22 n = 23 23(−7 + 103) ∴ S(23) = 2 1104 = (b) Let N be the number of terms that must be taken. ∵ a = –9, d = –2 – (–9) = 7 and S(n) = 1564 N and S(n) = [2a + (N – 1)d] 2 N ∴ 1564 = [2(–9) + (N – 1)(7)] 2 3128 = 7N2 – 25N 7N2 – 25N – 3128 = 0 (N – 23)(7N + 136) = 0 136 N = 23 or − (rejected) 7 ∴ 23 terms of the arithmetic series must be taken.

p. 183 1.

Arithmetic and Geometric Sequences and their Summation

(c) The required sum = sum of integers between 1 and 100 inclusive – sum of integers between 1 and 100 that are multiples of 3 = 5050 – 1683 (from (a) and (b)) = 3367

p. 195 1.

(a) ∵ a = 3, R = 2 and n = 6 3( 2 6 − 1) ∴ S(6) = 2 −1 = 189 (b) ∵ a = 39, R =

1 and n = 8 3

Certificate Mathematics in Action Full Solutions 5A

  1 8  39 1 −      3   ∴ S(8) = 1 1− 3 = 29 520 2.

N

1 1   < 3 19 683   1 1 < 3 N 39 3 N > 39 N >9 ∵ N is the least number and it is an integer. ∴ N = 10 ∴ There should be at least 10 terms.

(a) Let n be the number of terms of the given series. 512 1 = and T(n) = aRn – 1 = 8 ∵ a = 1024, R = 1024 2 ∴

1 8 = 1024   2

n −1

p. 202

n −1

1 1 =  128  2  n=8   1 8  1024 1 −      2   ∴ S(8) = 1 1− 2 = 2040 (b) Let N be the number of terms that must be taken. 1 a (1 − R N ) 7 ∵ a = 1024, R = and S(N) = = 2047 2 1− R 8   1 N  1024 1 −      2   7 2047 = ∴ 8 1 − 12

1.

2

− 1   1 (b) ∵ a = 1, R =  3  =− 1 3 1 3 S (∞ ) = =  1 4 ∴ 1− −   3 2.

N

1 1 2048  = 2 8   N = 14 ∴ 14 terms of arithmetic series must be taken. 3.

3

∴

N

  > 9841   N

19 682 1 1−   > 3 19 683   N

1 1 −  > − 3 19 683  

(a) 0.2 = 2222… = 0.2 + 0.02 + 0.002 + 0.0002 + … 0.2 = 1 − 0.1 2 = 9 (b) 0.2 4 = 0.242 424… = 0.24 + 0.0024 + 0.000 024 + … 0.24 = 1 − 0.01 8 = 33

Let the least number of terms of the geometric series be N. 37 1 a (1 − R N ) > 9841 ∵ a = 38, R = 8 = and S ( N ) = 3 3 1− R   1 N  38 1 −     3    > 9841 1 1− 39   1  1−   2   3 

1 (a) ∵ a = 1 and R =  2  1 = 1 2 1 S (∞ ) = =2 1 ∴ 1−

3.

1 C1B1 (mid-pt. theorem) 2 1 B2A2 = B1A1 (mid-pt. theorem) 2 1 A2C2 = A1C1(mid-pt. theorem) 2 ∴ Perimeter of △A2B2C2 1 = × perimeter of △A1B1C1 2 1 = (16 cm) 2

(a) C2B2 =

14

= 8 cm Similarly, perimeter of △A3B3C3 1 = × perimeter of △A2B2C2 2 1 = (8 cm) 2 = 4 cm

Arithmetic and Geometric Sequences and their Summation

T(2) = 2(2) + 3 = 7 T(3) = 2(3) + 3 = 9 T(4) = 2(4) + 3 = 11 2.

(b) Perimeter of △A2B2C2 1 = × perimeter of △A1B1C1 2 1 = × 16 cm 2 Perimeter of △A3B3C3 1 = × perimeter of △A2B2C2 2 1 1 = × × perimeter of △A1B1C1 2 2 1 1 = × × 16 cm 2 2

1 1 = 1 1 T(2) = 2 T(1) =

1 T(3) = 3 1 T(4) = 4 3.

T(1) = 12 – 3 = −2 T(2) = 22 – 3 = 1 T(3) = 32 – 3 = 6 T(4) = 42 – 3 = 13

2

1 =   × 16 cm 2 ∴ Perimeter of △AkBkCk 1 =  2 =2

5− k

4.

× 16 cm

5.

T(3) =

3(3) − 2 7 = 9 9

T(4) =

3( 4) − 2 10 = 9 9

T(1) = (–2)1 – 1 + 3 = 4 T(3) = (–2)3 – 1 + 3 = 7

p. 207

T(4) = (–2)4 – 1 + 3 = −5

X1 = 5, X2 = 7, X3 =

7+5 6+7 = 6, X4 = = 6.5, 2 2

6.

6.5 + 6 = 6.25 2

Y1 = 3, Y2 = 3(2 – 3) = –3, Y 3 = –3[2 – (–3)] = –15, Y 4 = –15[2 – (–15)] = –255, Y 5 = –255[2 – (–255)] = –65 535

Exercise Exercise 14A (p. 159) Level 1 1.

3( 2) − 2 4 = 9 9

T(2) = (–2)2 – 1 + 3 = 1

2

2.

T(2) =

cm

= 32 cm

X5 =

3(1) − 2 1 = 9 9

k −1

(c) From (b), the perimeters of the triangles formed are in 1 geometric sequence with common ratio . 2 16 = cm ∴ Sum of the perimeters 1 − 1

1.

T(1) =

T(1) = 2(1) + 3 = 5

80

T(1) = 32(1) – 1 = 3 T(2) = 32(2) – 1 = 27 T(3) = 32(3) – 1 = 243 T(4) = 32(4) – 1 = 2187

7.

(a) 64, 128 (b) ∵ T(1) = 4 = 21 + 1 T(2) = 8 = 22 + 1 T(3) = 16 = 23 + 1 T(4) = 32 = 24 + 1 n +1 ∴ T(n) = 2

Certificate Mathematics in Action Full Solutions 5A

(c) ∵ ∴ and ∴ 8.

are

9 11 and respectively. 10 12

Level 2

(a) 27, 33

11. T(1) =

31 1 37 311 = , T(7) = = 3 , T(11) = = 243 729 243 729 729

(b) ∵ T(1) = 3 = 6(1) – 3 T(2) = 9 = 6(2) – 3 T(3) = 15 = 6(3) – 3 T(4) = 21 = 6(4) – 3 ∴ T(n) = 6n − 3

12. T(1) =

2(1) 2 − 1 1 = 1+1 2

(c) ∵ ∴ and ∴ 9.

T(n) = 2n + 1 T(8) = 28 + 1 = 512 T(10) = 210 + 1 = 2048 The 8th term and the 10th term of the sequence are 512 and 2048 respectively.

T(n) = 6n – 3 T(8) = 6(8) – 3 = 45 T(10) = 6(10) – 3 = 57 The 8th term and the 10th term of the sequence are 45 and 57 respectively.

(a) –1, 1 (b) ∵ T(1) = –1 = (–1)1 T(2) = 1 = (–1)2 T(3) = –1 = (–1)3 T(4) = 1 = (–1)4 n ∴ T(n) = (−1) n

(c) ∵ T(n) = (–1) ∴ T(8) = (–1)8 = 1 and T(10) = (–1)10 = 1 ∴ The 8th term and the 10th term of the sequence are 1 and 1 respectively. 10. (a)

6 7 , 7 8 2 1+1 = 3 1+ 2 3 2 +1 T(2) = = 4 2+2 4 3 +1 T(3) = = 5 3+ 2 5 4 +1 T(4) = = 6 4+2 n +1 ∴ T(n) = n+2

(b) ∵ T(1) =

n +1 n+2 8 +1 9 = ∴ T(8) = 8 + 2 10 10 + 1 11 = and T(10) = 10 + 2 12 ∴ The 8th term and the 10th term of the sequence

(c) ∵ T(n) =

T(7) =

2(7) 2 − 1 97 = 7 +1 8

T(11) =

2(11) 2 − 1 241 = 11 + 1 12

(1 − 1)(1 − 2) 0 = 1 (7 − 1)(7 − 2) 30 = T(7) = 7 7

13. T(1) =

T(11) =

(11 − 1)(11 − 2) 90 = 11 11

31 3 = 13 37 2187 T(7) = 3 = 7 343

14. T(1) =

T(11) =

311 177 147 = 113 1331

15. (a) –5, 6 (b) ∵ T(1) = –1 = (–1)1 × 1 T(2) = 2 = (–1)2 × 2 T(3) = –3 = (–1)3 × 3 T(4) = 4 = (–1)4 × 4 n ∴ T(n) = (−1) n

16. (a) log 80, log 160 (b) ∵ T(1) = log 5 = log (5 ⋅ 21 – 1) T(2) = log 10 = log (5 ⋅ 22 – 1) T(3) = log 20 = log (5 ⋅ 23 – 1) T(4) = log 40 = log (5 ⋅ 24 – 1) n −1 ∴ T(n) = log (5 ⋅ 2 ) 17. (a) 30, 42 (b) ∵ T(1) = 2 = 1 × (1 + 1) T(2) = 6 = 2 × (2 + 1)

14

T(3) = 12 = 3 × (3 + 1) T(4) = 20 = 4 × (4 + 1) ∴ T(n) = n( n + 1) 18. (a)

Arithmetic and Geometric Sequences and their Summation

(b) T(10) = 3 + 2(10) = 23 6.

25 36 , 36 49

(a) ∵ a = –3 and d = 9 and T(n) = a + (n – 1)d ∴ T(n) = –3 + (n – 1)(9) = −12 + 9n (b) T(10) = –12 + 9(10) = 78

2

(b) ∵ T(1) =

1 1 = 4 (1 + 1) 2

T(2) =

4 22 = 9 (2 + 1) 2

T(3) =

9 32 = 16 (3 + 1) 2

T(4) =

16 42 = 25 ( 4 + 1) 2

∴ T(n) =

n2 (n + 1) 2

7.

3 and d = –2 2 and T(n) = a + (n – 1)d 7 3 ∴ T(n) = + (n – 1)(–2) = − 2n 2 2

(a) ∵ a =

(b) T(10) =

8.

Exercise 14B (p. 166)

33 7 – 2(10) = − 2 2

3 2 and T(n) = a + (n – 1)d

(a) ∵ a = –2 and d =

∴ T(n) = –2+ (n – 1)(

Level 1 1.

T(2) – T(1) = 12 – 16 = –4 T(3) – T(2) = 8 – 12 = –4 T(4) – T(3) = 4 – 8 = –4 ∴ It is an arithmetic sequence with common difference – 4.

(b) T(10) =

9. 2.

T(2) – T(1) = 2 − 1 = 2 − 1 T(3) – T(2) = 3 − 2 ≠ 2 − 1 ∴ It is not an arithmetic sequence.

3.

5 1 1 T(2) – T(1) = − = 6 3 2 4 5 1 T(3) – T(2) = − = 3 6 2 11 4 1 T(4) – T(3) = − = 6 3 2 ∴ It is an arithmetic sequence with common difference 1 . 2 2 1 1 − = 3 2 6 3 2 1 1 ≠ T(3) – T(2) = − = 4 3 12 6 ∴ It is not an arithmetic sequence.

4.

T(2) – T(1) =

5.

(a) ∵ a = 5 and d = 2 and T(n) = a + (n – 1)d ∴ T(n) = 5 + (n – 1)(2) = 3 + 2n

80

1 3 ) = (3n − 7) 2 2

23 1 [3(10) – 7] = 2 2

d = 6 – 1 =5 T(n) = a + (n – 1)d = 1 + (n – 1)(5) = 5n − 4

10. d = 17 – 19 = −2 T(n) = a + (n – 1)d = 19 + (n – 1)(–2) = 21− 2n 11. d = log 9 – log 3 = 2 log 3 – log 3 = log 3 T(n) = a + (n – 1)d = log 3 + (n – 1)(log 3) = n log 3 12. d = (2a – 3c) – (a – 2c) = a − c T(n) = a + (n – 1)d = (a – 2c) + (n – 1)(a – c) = −c + n ( a − c ) 13. Let a and d be the first term and the common difference respectively. ∵ a = 7 and d = 11 – 7 = 4 ∴ T(n) = a + (n – 1)d = 7 + (n – 1)(4) = 3 + 4n Let 83 be the kth term. i.e. T(k) = 83 ∴ 3 + 4k = 83 k = 20 ∴ There are 20 terms in the sequence.

Certificate Mathematics in Action Full Solutions 5A

14. Let a and d be the first term and the common difference respectively. ∵ a = –93 and d = –86 – (–93) = 7 ∴ T(n) = a + (n – 1)d = –93 + (n – 1)(7) = –100 + 7n Let 5 be the kth term. i.e. T(k) = 5 ∴ –100 + 7k = 5 k = 15 ∴ There are 15 terms in the sequence. 15. Let a and d be the first term and the common difference respectively. T(2) = a + d = 18 ……(1) T(6) = a + 5d = 30 ……(2) (2) – (1), 4d = 12 d=3 By substituting d = 3 into (1), we have a + 3 = 18 a = 15 ∴ T(n) = a + (n – 1)d = 15 + (n – 1)(3) = 12 + 3n 16. Let a and d be the first term and the common difference respectively. T(4) = a + 3d = –2 ……(1) T(9) = a + 8d = –32 ……(2) (2) – (1), 5d = –30 d = –6 By substituting d = –6 into (1), we have a + 3(–6) = –2 a = 16 ∴ T(n) = a + (n – 1)d = 16 + (n – 1)(–6) = 22 − 6n 17. Let a and d be the first term and the common difference respectively. T(3) = a + 2d = 60 ……(1) T(7) = a + 6d = 40 ……(2) (2) – (1), 4d = –20 d = –5 By substituting d = –5 into (1), we have a + 2(–5) = 60 a = 70 ∴ T(n) = a + (n – 1)d = 70 + (n – 1)(–5) = 75 − 5n 18. Let a and d be the first term and the common difference respectively. T(3) = a + 2d = 82 ……(1) T(10) = a + 9d = 250 ……(2) (2) – (1), 7d = 168 d = 24 By substituting d = 24 into (1), we have a + 2(24) = 82 a = 34 ∴ T(n) = a + (n – 1)d = 34 + (n – 1)(24) = 10 + 24n 19. (a) 1, 2; 2, 4; 3, 6 (or any other reasonable answers) (b) –1, 1; –2, 2; –3, 3 (or any other reasonable answers)

Level 2 20. Let a and d be the first term and the common difference respectively. a = –101,d = –98 – (–101) = 3 ∴ T(n) = a + (n – 1)d = –101 + (n – 1)(3) = –104 + 3n ∵ The kth term is the first positive term of the sequence. ∴ T(k) > 0 i.e. –104 + 3k > 0 104 k> 3 ∵ k is the number of terms, it must be an integer. ∴ k = 35 21. Let a and d be the first term and the common difference respectively. a = 999, d = 992 – 999 = –7 ∴ T(n) = a + (n – 1)d = 999 + (n – 1)(–7) = 1006 – 7n Let the kth term be the last positive term. ∴ T(k) > 0 i.e. 1006 – 7k > 0 1006 k< 7 ∵ k is the number of terms, it must be an integer. ∴ k = 143 ∴ There are 143 positive terms. 22. Let a, (a + d) and (a + 2d) be the interior angles of the triangle respectively. ∵ a = 15° ∴ The other two angles are 15° + d and 15° + 2d. 15° + (15° + d) + (15° + 2d) = 180° (∠s sum of triangle) d = 45° ∴ The largest angle is 105°. 23. The multiples of 13 form an arithmetic sequence with a = 13 and d = 13. ∴ T(n) = a + (n – 1)d = 13 + (n – 1)(13) = 13n Consider the term that is less than 1000. i.e. T(n) < 1000 13n < 1000 1000 n< 13 ∵ Among the terms that is less than 1000, the 76th term is the greatest. ∴ The greatest integer which is a multiple of 13 and less than 1000 is 13(76) = 988. 24. (a) Let a and d be the first term and the common difference respectively. T(5) = 4T(1) a + 4d = 4a 3a – 4d = 0 ……(1)

14

T(6) = 2T(3) – 1 a + 5d = 2(a + 2d) – 1 a–d=1 ……(2) (2) × 4 – (1), a = 4 By substituting a = 4 into (2), we have 4–d=1 d=3 ∴ T(n) = a + (n – 1)d = 4 + (n – 1)(3) = 1+ 3n (b)

T(m) – T(25) = 81 [1 + 3m] – [1 + 3(25)] = 81 m = 52

25. Let (a – d) cm, a cm and (a + d) cm be the lengths of the sides of the right-angled triangle respectively. ∵ The perimeter is 27 cm. ∴ (a – d) + a + (a + d) = 27 a=9 (a – d)2 + a2 = (a + d)2 (Pyth. theorem) a2 – 2ad + d2 + a2 = a2 + 2ad + d2 a2 = 4ad ( a ≠ 0) a = 4d 1 d = a………(1) 4 By substituting a = 9 into (1), we have d = 2.25. ∴ The lengths of the three sides of the triangle are 6.75 cm, 9 cm and 11.25 cm.

When d = 3, a – d = 5 – 3 = 2 and a + d = 5 + 3 = 8 ∴ The terms are 2, 5, 8. 28. Let a and d be the first term and the common difference respectively. aT(6) = –38 a(a + 5d) = –38 ………(1) T(3) + T(4) = 17 (a + 2d) + (a + 3d) = 17 17 − 2a d= ………(2) 5 By substituting (2) into (1), we have  17 − 2a   ] = –38 a[a + 5   5  a2 – 17a – 38 = 0 (a – 19)(a + 2) = 0 a = 19 or a = –2(rejected) By substituting a = 19 into (2), we have 17 − 2(19) 21 =− d= 5 5 ∴ The first term and the common difference are 19 and 21 − respectively. 5 29. (a) T(2) – T(1) = log 10k2 – log 10k 10k 2 10k = log k T(3) – T(2) = log 10k3 – log 10k2

26. (a) ∵ T(9) = 3T(4) ∴ a + 8d = 3(a + 3d) a + 8d = 3a + 9d d a=− 2

= log

d + (n – 1)d 2  3  =  − 2 + nd  

∴ T(n) = a + (n – 1)d = −

(b) ∵ T(k) = 5T(5) 3    3   − + k d = 5 − + 5 d  2   2  k = 19 27. Let the three terms be a – d, a and a + d respectively. ∵ The sum is 15. ∴ (a – d) + a + (a + d) = 15 a=5 ∵ The product is 80. ∴ (a – d)(a)(a + d) = 80 ………(1) By substituting a = 5 into (1), we have (5 – d)(5)(5 + d) = 80 d = ±3 When d = –3, a – d = 5 – (–3) = 8 and a + d = 5 + (–3) = 2

80

Arithmetic and Geometric Sequences and their Summation

10k 3 10k 2 = log k T(4) – T(3) = log 10k4 – log 10k3 = log

10k 4 10k 3 = log k It is an arithmetic sequence. a = log 10k and d = log k T(n) = a + (n – 1)d = log 10k + (n – 1)(log k) = 1 + n log k The general term is 1 + n log k. = log

∴ ∵ ∴

∴

(b) T(2) – T(1) = log (10k2)2 – log (10k)2 = 2(log 10k2 – log 10k) = 2 log k T(3) – T(2) = log (10k3)2 – log (10k2)2 = 2(log 10k3 – log 10k2) = 2 log k T(4) – T(3) = log (10k4)2 – log (10k3)2 = 2(log 10k4 – log 10k3) = 2 log k

Certificate Mathematics in Action Full Solutions 5A

∴ It is an arithmetic sequence. ∵ a = log (10k)2 and d = 2 log k ∴ T(n) = a + (n – 1)d = log (10k)2 + (n – 1)(2 log k) = 2 + 2n log k ∴ The general term is 2 + 2n log k. 30. (a) Let the three terms be a , a + d and a + 2d respectively. a(a + d)(a + 2d) = a + (a + d) + (a + 2d) a(a + d)(a + 2d) = 3(a + d) [a(a + 2d) – 3] (a + d) = 0 a(a + 2d) – 3 = 0 or a = −d a2 + 2ad – 3 = 0 a=

− 2d ± (2d ) 2 − 4(−3) 2(1)

Exercise 14C (p. 170) Level 1 1.

(b) Arithmetic mean =

7 + 25 = 16 2

(c) Arithmetic mean =

42 + 104 = 73 2

8+ x = 25 2 x = 42

∵

3.

(a) Let d1 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d1, –4 + 2d1, 8 ∵ The 4th term is also given by –4 + 3d1. ∴ –4 + 3d1 = 8 d1 = 4 ∴ The two required arithmetic means are 0 and 4.

(b) If a = − d + d 2 + 3 , then a = − 1 + 12 + 3 = 1. ∴ The terms are 1, 2, 3. If a = − d − d 2 + 3 , then a = − 1 − 12 + 3 = –3.

(b) Let d2 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d2, –4 + 2d2, –4 + 3d2, 8 ∵ The 5th term is also given by –4 + 4d2. ∴ –4 + 4d2 = 8 d2 = 3 ∴ The three required arithmetic means are –1, 2 and 5.

∴ The terms are –3, –2, –1. If a = –d, then a = –1. ∴ The terms are –1, 0, 1. 31. Let the three integers be a – d, a and a + d respectively, and their sum be k. (a – d) + a + (a + d) = k 3a = k………(1) (a – d)(a)(a + d) = 11k………(2) By substituting (1) into (2), we have (a – d)(a)(a + d) = 11(3a) ( a ≠ 0) (a – d)(a + d) = 33

(c) Let d3 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d3, –4 + 2d3, –4 + 3d3, –4 + 4d3, –4 + 5d3, –4 + 6d3, –4 + 7d3, 8 ∵ The 9th term is also given by –4 + 8d3. ∴ –4 + 8d3 = 8 d3 = 1.5 ∴ The seven required arithmetic means are –2.5, – 1, 0.5, 2, 3.5, 5 and 6.5.

∵ a and d are integers. ∴ Possible solutions are:

a = 17 a = 7 a = 7 a = 17 i.e.  or  or  or  d = 16 d = 4 d = −4 d = −16 If a = 17 and d = 16, the three integers are 1, 17, 33. If a = 7 and d = 4, the three integers are 3, 7, 11. If a = 7 and d = –4, the three integers are 11, 7, 3. If a = 17 and d = –16, the three integers are 33, 17, 1. ∴ The three integers are 1, 17, 33 or 3, 7, 11.

−8 + 4 = −2 2

2.

a =−d ± d2 +3

a + d = 33 a + d = 11 a + d = 3 or  or  or  a − d = 1 a − d = 3 a − d = 11 a + d = 1  a − d = 33

(a) Arithmetic mean =

4.

Let d be the common difference of the arithmetic sequence. ∵ a = 18 and T(5) = 36 ∴ 36 = 18 + 4d 9 d= 2 9 ∴ a = 18 +   = 22.5 2 9 b = 18 + 2   = 27 2

14

Arithmetic and Geometric Sequences and their Summation

9 c = 18 + 3   = 31.5 2 5.

arithmetic mean = 13

1, 2, 4, 5; –1, 1, 5, 7; –3, 0, 6, 9 (or any other reasonable answers)

Level 2 6.

7.

8.

(a) Let d be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 1, 1 + d, 1 + 2d, 1 + 3d, p ∵ The 5th term is also given by 1 + 4d. ∴ 1 + 4d = p p −1 d= 4 ∴ The three required arithmetic means are p + 3 p +1 3p +1 , and . 4 2 4 (b) common difference =

(ii) ∵ The (n +2)th term is also given by x + (n +1)d1. ∴ x + (n +1)d1 = y y−x d1 = n +1 (b) (i) The arithmetic sequence formed is: x, x + d2, x + 2d2, x + 3d2, …, x + md2, y ∴ The arithmetic means are x + d2, x + 2d2, x + 3d2, …, x + md2 (ii) ∵ The (m +2)th term is also given by x + (m +1)d2. ∴ x + (m +1)d2 = y y−x d2 = m +1 (c) d1 : d2 =

( a − d ) + (a + d ) 2a = =a 2 2 (b) By substituting a = 132 and d = 92 into (a), we have

Level 1 1.

2.

3.

(a) Arithmetic mean =

80

y−x y−x : = (m + 1) : ( n + 1) n +1 m +1

Exercise 14D (p. 176)

p −1 4 (from (a))

 p −1  T(n) = a + (n – 1)d = 1 + (n – 1)   4  (n − 1)( p − 1) =1+ 4 ( n − 1)( p − 1) ∴ The general term is 1 + . 4 9.

10. (a) (i) The arithmetic sequence formed is: x, x + d1, x + 2d1, x + 3d1, …, x + nd1, y ∴ The arithmetic means are x + d1, x + 2d1, x + 3d1, …, x + nd1

∵ 2x + 1 is the arithmetic mean between x and 14. x + 14 ∴ 2x + 1 = 2 4x + 2 = 14 + x x=4 ∴ a = 4 and d = (2x + 1) – x = x + 1 = 5 T(n) = a + (n – 1)d = 4 + (n – 1)(5) = 5n − 1 ∵ x is the arithmetic mean between 8 and y, and 21 is the arithmetic mean between y and 26. 8 + y .........(1)   x= 2 ∴ y + 26 21 = .........(2) 2  From (2), we have 42 = y + 26 y = 16 By substituting y = 16 into (1), we have 8 + 16 = 12 x= 2

2

4.

T (2) 10 = =2 T (1) 5 T (3) 15 3 = = ≠2 T (2) 10 2 ∴ It is not a geometric sequence. T (2) −8 = = –4 T (1) 2 T (3) 32 = = –4 T (2) − 8 T (4) −128 = = –4 T (3) 32 ∴ It is a geometric sequence with common ratio –4. T (2) −10 = = –5 T (1) 2 T (3) −50 = = 5 ≠ −5 T (2) − 10 ∴ It is not a geometric sequence. T (2) 0.1 = = 0.5 T (1) 0.2

Certificate Mathematics in Action Full Solutions 5A

T (3) 0.05 = = 0.5 T (2) 0.1 T (4) 0.025 = = 0.5 T (3) 0.5 ∴ It is a geometric sequence with common ratio 0.5. 5.

(a) ∵ a = 2 and R = 3 and T(n) = aRn – 1 n−1 ∴ T(n) = 2 ⋅ 3

∴ T(n)

1 = (−3)  3 2− n = −3

n −1

(b) T(8) = –32 – 8 1  1 = − 729  or − 36    7.

(a) ∵ a = –4 and R = −  3 ∴ T(n) = ( −4) −   2  3 (b) T(8) = ( −4) − 2   

8.

8−1

=

3 and T(n) = aRn – 1 2 n −1

(or ( −1) n ⋅ 23−n ⋅ 3n−1 )

2187  37   or  32  25 

(a) ∵ a = 2 and R = − 2 and T(n) = aRn – 1 ∴ T(n) = 2 (− 2 ) n−1 n   = ( −1) n−1 ( 2 ) n  or (−1) n−1 ⋅ 2 2   

8−1 8 (b) T(8) = ( −1) ( 2 ) = − 16

9.

Let a be the first term. 6 ∵ a = 2 and R = = 3 2 ∴ T(n) = aRn – 1 n −1 = 2⋅3

10. Let a be the first term. 2 1 ∵ a = 6 and R = = 6 3 ∴ T(n) = aRn – 1

11. Let a be the first term. ∵ a = 2 2 and R =  3   = 2 2 2   

1 and T(n) = aRn – 1 3

(a) ∵ a = –3 and R =

n −1

2 3 2 2

=

3 2

∴ T(n) = aRn – 1

(b) T(8) = 2 ⋅ 38−1 7 = 4374 (or 2 ⋅ 3 ) 6.

1 = 6  3 2 = n−2 3

n −1

 4−n n−1  or  2 2 ⋅ 3 2   

12. Let k be the first term. a2r ∵ k = a and R = = ar ar n–1 ∴ T(n) = kR n n−1 = a ⋅r 13. Let a and R be the first term and the common ratio respectively. 243 1 = a = 729, R = 729 3 n–1 ∴ T(n) = aR n −1

1 = 729   3 = 37 – n Let 1 be the kth term. i.e. T(k) = 1 ∴ 37 – k = 1 ∴ 7–k=0 k=7 ∴ There are 7 terms in the sequence. 14. Let a and R be the first term and the common ratio respectively. −3 a = 1.5, R = = –2 1.5 n–1 ∴ T(n) = aR = 1.5(–2)n – 1 Let 96 be the kth term. i.e. T(k) = 96 ∴ 1.5(–2)k – 1 = 96 (–2)k – 1 = 64 (–2)k – 1 = (–2)6 ∴ k–1=6 k=7 ∴ There are 7 terms in the sequence. 15. Let a and R be the first term and the common ratio respectively. 1 T(2) = aR = ………(1) 4

14

T(7) = aR6 = 8 (2) ÷ (1),

………(2) R5 = 32 R=2 By substituting R = 2 into (1), we have 1 a(2) = 4 1 a= 8 ∴ T(n) = aRn – 1 1 = ⋅ 2n – 1 8 =2

n −4

16. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = –27 ………(1) T(6) = aR5 = 1 ………(2) 1 (2) ÷ (1), R3 = − 27 1 R=− 3 1 By substituting R = − into (1), we have 3 2

 1 a  −  = –27  3 a = –243 ∴ T(n) = aRn – 1 n −1  1 = − 243 −   3 n ( − 1 ) ⋅ 36− n = 17. Let a and R be the first term and the common ratio respectively. T(2) = aR = 5 ………(1) T(7) = aR6 = 160 ………(2) (2) ÷ (1), R5 = 32 R=2 By substituting R = 2 into (1), we have a(2) = 5 5 a= 2 ∴ T(n) = aRn – 1 5 = ⋅ 2n – 1 2 n−2 5 = ⋅2 18. Let a and R be the first term and the common ratio respectively. T(2) = aR = 243 ………(1) 1 T(5) = aR4 = ………(2) 3

80

Arithmetic and Geometric Sequences and their Summation

1 729 1 R= 9 1 By substituting R = into (1), we have 9 1 a   = 243 9 a = 2187 ∴ T(n) = aRn – 1 n −1 1 = 2187 ⋅   9 = 39−2 n (2) ÷ (1),

R3 =

19. (a) 2, 4; 3, 9; 4, 16 (or any other reasonable answers) (b)

1 1 1 , 3; , 2; − , –3 3 2 3 (or any other reasonable answers)

Level 2 20. Let a and R be the first term and the common ratio respectively. T(2) = aR = 6 ………(1) T(6) = aR5 = 96 ………(2) (2) ÷ (1), R4 = 16 R = ±2 By substituting R = 2 into (1), we have a(2) = 6 a=3 ∴ T(n) = aRn – 1 = 3 ⋅ 2n–1 By substituting R = –2 into (1), we have a(–2) = 6 a = –3 ∴ T(n) = aRn – 1 = –3(–2)n – 1 ∴ The general term is 3 ⋅ 2n – 1 or –3(–2)n – 1. 21. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = 1 ………(1) 1 6 T(7) = aR = ………(2) 16 1 (2) ÷ (1), R4 = 16 1 R=± 2 1 By substituting R = into (1), we have 2 2

1 a  = 1 2 a=4

Certificate Mathematics in Action Full Solutions 5A

(2) ÷ (1),

∴ T(n) = aRn – 1 1 = 4  2 = 23 – n

n −1

R=± 6 By substituting R = 6 into (1), we have

1 By substituting R = − into (1), we have 2 2

 1 a −  = 1  2 a=4 ∴ T(n) = aRn – 1 n −1

22. Let a and R be the first term and the common ratio respectively. T(2) = aR = –9 ………(1) 729 T(6) = aR5 = − ………(2) 256 81 (2) ÷ (1), R4 = 256 3 R=± 4 3 By substituting R = into (1), we have 4 3 a   = –9 4 a = –12 ∴ T(n) = aRn – 1 3 = (–12)   4 By substituting R = −

n −1

3 into (1), we have 4

 3 a  −  = –9  4 a = 12 ∴ T(n) = aRn – 1  3 = 12  −   4

n −1

3 ∴ The general term is (–12)   4 n −1

n −1

(or –24–2n ⋅ 3n) or 12

(or (–1)n–1 ⋅ 24 – 2n ⋅ 3n).

23. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = 6 3 ………(1) T(5) = aR4 = 36 3

a( 6 ) 2 = 6 3 a= 3 ∴ T(n) = aRn – 1 = 3 ( 6 ) n−1 By substituting R = − 6 into (1), we have

 1 = 4 −   2 = (–1)n – 1 ⋅ 23 – n ∴ The general term is 23 – n or (–1)n – 1 ⋅ 23 – n.

 3 −   4

R2 = 6

………(2)

a (− 6 ) 2 = 6 3 a= 3 ∴ T(n) = aRn – 1 = 3 ( − 6 ) n−1 n −1 n   2 2 3 ( 6 ) n−1  or 2 ⋅ 3    or ∴ The general term is n −1 n   n−1 2 2 3 ( − 6 ) n−1  or (−1) ⋅ 2 ⋅ 3  .  

24. Let n be the number of years taken. Consider the salary of Leo, a = $10 000, R = 1 + 10.25% = 1.1025 ∴ The salary of Leo after n years = $10 000 × 1.1025n Consider the salary of Michael, a = $12 000, R = 1 + 5% = 1.05 ∴ The salary of Michael after n years = $12 000 × 1.05n For Leo to have a salary higher than that of Michael, we have $10 000 × 1.1025n > $12 000 × 1.05n 5 × 1.1025n > 6 × 1.05n log (5 × 1.1025n) > log (6 × 1.05n) log 5 + n log 1.1025 > log 6 + n log 1.05 n (log 1.1025 – log 1.05) > log 6 – log 5 log 6 − log 5 n> log1.1025 − log1.05 n > 3.7 ∴ The number of years taken is 4. 25. Let A be the surface area of the pond and a be the area covered by the lotus leaves after 1 week. ∵ R=2 and after 16 weeks, area covered = A ∴ a(216 – 1) = A A a = 15 2 Let it takes k weeks to cover one eighth of the pond. A ∴ a(2k – 1) = 8 A k −1 A (2 ) = 8 215 2k – 1= 212

14

k = 13 ∴ It takes 13 weeks to cover one eighth of the pond. 26. (a) Let the vertices of the kth triangle be Ak, Bk and Ck, k = 1, 2, 3,…

h = 12.5 ∴ The height of the building is 12.5 m. 28. (a) At the end of 2001, population = 5 000 000 × (1 + 4%) = 5 200 000 At the end of 2002, population = 5 200 000 × (1 + 4%) = 5 000 000 × (1 + 4%)2 = 5 408 000 (b) Let the population will be doubled at the end of the nth year. 5 000 000 × (1 + 4%)n > 2 × 5 000 000 n log 1.04 > log 2 n > 17.7 ∴ At the end of 2018, the population will be doubled.

1 C1B1 (mid-pt. theorem) 2 1 B2A2 = B1A1 (mid-pt. theorem) 2 1 A2C2 = A1C1(mid-point theorem) 2 ∴ Perimeter of second triangle = C2B2 + B2A2 + A2C2 1 1 1 = C1B1 + B1A1 + A1C1 2 2 2 1 = × perimeter of first triangle 2 Perimeter of second traingle 1 = ∴ Perimeter of first traingle 2 C 2B 2 =

29. Let the original numbers be k, 5k and 11k respectively. ∵ k + 3, 5k + 3, 11k + 3 is a geometric sequence. 11k + 3 5k + 3 = ∴ 5k + 3 k +3 2 (5k + 3) = (11k + 3)(k + 3) 25k2 + 30k + 9 = 11k2 + 36k + 9 14k2 – 6k = 0 k(7k – 3) = 0 3 k = or k = 0(rejected) 7 3 15 33 ∴ The original numbers are , and . 7 7 7

(b) By an argument similar to (a), Perimeter of third triangle 1 = × perimeter of second triangle 2 1 1 = × × perimeter of first triangle 2 2

Exercise 14E (p. 181) Level 1 1.

(c) Geometric mean = ± − 4 × (−36) = 12 or − 12

4

1 ∴ Perimeter of fifth triangle =   x 2

2.

∵ The geometric mean between x and 48 is 12. ∴ 122 = 48x x =3

3.

(a) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 243, 243R, 243R2, 9 ∵ The 4th term is also given by 243R3. ∴ 243R3 = 9 1 R= 3 ∴ The two required geometric means are 81 and 27.

4

1 1.5 =   x 2 x = 24

27. Let the height of the building be h m. 3 ∵ R= 5 4

 3 ∴ After the 4th rebound, the height =   h 5 4

 3 1.62 =   h 5

i.e.

80

(a) Geometric mean = 9 × 27 = 9 3 (b) Geometric mean = − 4 × 16 = − 8

2

1 =  x 2

i.e.

Arithmetic and Geometric Sequences and their Summation

Certificate Mathematics in Action Full Solutions 5A

(b) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: –16, –16r, –16r2, –16r3, –81 ∵ The 5th term is also given by –16r4. ∴ –16r4 = –81 3 r=± 2 ∴ The three required geometric means are 24, –36 and 54 or –24, –36 and –54.

(b) ∵ The geometric mean between a and 2 is 3 2 . ∴

8.

∵ m, –2, n are in geometric sequence. ∴ –2 is the geometric mean between m and n. ∴ (–2)2 = mn mn = 4 ……(1) ∵ –2, n, m are in arithmetic sequence. ∴ n is the arithmetic mean between –2 and m. ∴ 2n = m + (–2) m = 2n + 2 ……(2) By substituting (2) into (1), we have (2n + 2)n = 4 n2 + n – 2 = 0 (n – 1)(n + 2) = 0 n = 1 or n = –2 (rejected) By substituting n = 1 into (2), we have m = 2(1) + 2 = 4

9.

(a) b = a + a tan θ = a(1 + tan θ) c = b + b tan θ = b(1 + tan θ) ac = a[b(1 + tan θ)] = b[a(1 + tan θ)] = b2 ∴ b is a geometric mean between a and c.

(c) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 1, r, r2, r3, r4, 32 ∵ The 6th term is also given by r5. ∴ r5 = 32 r=2 ∴ The four required geometric means are 2, 4, 8 and 16. 4.

5.

Let R be the common ratio. The geometric sequence is: 54, 54R, 54R2, 2 ∵ The 4th term is also given by 54R3. ∴ 54R3 = 2 1 R= 3 ∴ The value of a and b are 18 and 6 respectively.

b a (1 + tan θ ) = = 1 + tan θ a a ∴ The common ratio is 1 + tan θ. If θ = 30°, 1 + tan θ = 1 + tan 30° 1 +1 = 3

(b) ∵

2, 18 or 3, 12 or –2, –18 or –3, –12 (or any other reasonable answers)

Level 2 6.

7.

∵ x + 1, x + 5, 2x + 4 are in geometric sequence. ∴ x + 5 is the geometric mean between x + 1 and 2x + 4. ∴ (x + 5)2 = (x +1)(2x + 4) 2 x + 10x + 25 = 2x2 + 6x + 4 x2 – 4x – 21 = 0 (x + 3) (x – 7) = 0 x = −3 or 7 (a) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: a2 a2 a2 2 4 , R, R , 2 2 2 a ∵ The 4th term is also given by

a2 3 R . 2

a2 3 4 R = ∴ 2 a 2 R= a ∴ The two geometric means are a and 2.

(3 2 ) 2 = 2a a =9

Exercise 14F (p. 189) Level 1 1.

(a) ∵ a = 1, d = 3 – 1 = 2 and n = 20 20 ∴ S(20) = [2(1) + (20 – 1)(2)] 2 = 400 (b) ∵ a = –5, d = –2 – (–5) = 3 and n = 25 25 ∴ S(25) = [2(–5) + (25 – 1)(3)] 2 = 775 (c) ∵ a = 78, d = 72 – 78 = –6 and n = 27 27 ∴ S(27) = [2(78) + (27 – 1)(–6)] 2 =0 (d) ∵ first term = (a + b), d = (3a – b) – (a + b) = 2a – 2b and n = 10

14

10 [2(a + b) + (10 – 1)(2a – 2b)] 2 = 100a − 80b

∴ S(10) =

2.

(d) Let a and d be the first term and the common difference respectively. 1 1 1 1 ∵ a = and d = − = 3 2 3 6 ∴ T(n) = a + (n – 1)d 1 1 = 3 + ( n − 1) 6 1 1 =6+6n 5 Let 1 be the kth term. 6 5 i.e. T(k) = 1 6 1 1 5 + k =1 ∴ 6 6 6 k = 10 ∴ There are 10 terms in the sequence. 5 1 10 + 1  6 ∴ S(10) =  3 2 5 = 10 6

(a) Let a and d be the first term and the common difference respectively. ∵ a = 1 and d = 9 – 1 = 8 ∴ T(n) = a + (n – 1)d = 1 + (n – 1)(8) = 8n – 7 Let 97 be the kth term. i.e. T(k) = 97 ∴ 8k – 7 = 97 k = 13 ∴ There are 13 terms in the sequence. 13(1 + 97) ∴ S(13) = 2 = 637 (b) Let a and d be the first term and the common difference respectively. ∵ a = –1 and d = 2 – (–1) = 3 ∴ T(n) = a + (n – 1)d = –1 + (n – 1)(3) = 3n – 4 Let 95 be the kth term. i.e. T(k) = 95 ∴ 3k – 4 = 95 k = 33 ∴ There are 33 terms in the sequence. 33( −1 + 95) ∴ S(33) = 2 = 1551 (c) Let a and d be the first term and the common difference respectively. 3 1 3 3 ∵ a = 5 and d = 6 − 5 = 4 2 4 4 ∴ T(n) = a + (n – 1)d 3 3 = 5 4 + ( n − 1) 4 4 =5 + 3 n Let 32 be the kth term. i.e. T(k) = 32 3 ∴ 5 + k = 32 4 k = 36 ∴ There are 36 terms in the sequence.  3  36 5 + 32   ∴ S(36) =  4 2 1 = 679 2

80

Arithmetic and Geometric Sequences and their Summation

3.

(a) Let N be the number of terms that must be taken. ∵ a = 7, d = 10 – 7 = 3 and S(n) = 920 N and S(n) = [2a + (N – 1)d] 2 N ∴ 920 = [2(7) + (N – 1)(3)] 2 1840 = 3N2 – 11N 3N2 – 11N – 1840 = 0 (N – 23)(3N + 80) = 0 80 N = 23 or − (rejected) 3 ∴ 23 terms of arithmetic series must be taken. (b) Let N be the number of terms that must be taken. ∵ a = 21, d = 15 – 21 = –6 and S(n) = –60 N and S(n) = [2a + (N – 1)d] 2 N ∴ –60 = [2(21) + (N – 1)(–6)] 2 –120 = –6N2 + 48N N2 – 8N – 20 = 0 (N – 10)(N + 2) = 0 N = 10 or –2(rejected) ∴ 10 terms of arithmetic series must be taken. (c) Let N be the number of terms that must be taken. ∵ a = 27, d = 24 – 27 = –3 and S(n) = 126 N and S(n) = [2a + (N – 1)d] 2

Certificate Mathematics in Action Full Solutions 5A

N [2(27) + (N – 1)(–3)] 2 252 = –3N2 + 57N N2 – 19N + 84 = 0 (N – 7)(N – 12) = 0 N = 7 or 12 ∴ 7 or 12 terms of arithmetic series must be taken. 126 =

∴

7.

∵ a = 100, d = –10 7 ∴ S(7) = $ [2(100) + (7 – 1)(–10)] 2 = $490 > $480 ∴ She has enough money to buy the watch.

8.

(a) ∵ a = 1, l = 200 and n = 200 200(1 + 200) ∴ 1 + 2 + … + 200 = 2 = 20100

(d) Let N be the number of terms that must be taken. 1 4 ∵ a = 6, d = 7 − 6 = and S(n) = 2470 3 3 N and S(n) = [2a + (N – 1)d] 2 4 N ∴ 2470 = [2(6) + (N – 1)   ] 2 3 14 820 = 4N2 + 32N N2 + 8N – 3705 = 0 (N – 57)(N + 65) = 0 N = 57 or –65(rejected) ∴ 57 terms of arithmetic series must be taken. 4.

5.

∵ d = 2 and T(4) = 10 ∴ a + 3(2) = 10 a=4 10 ∴ S(10) = [2(4) + (10 – 1)2] 2 = 130 (a) ∵

S(14) = 406 14 ∴ [2(a) + (14 – 1)d] = 406 2 2a +13d = 58 ………(1) ∵ T(4) + T(5) = 34 ∴ (a + 3d) + (a + 4d) = 34 2a +7d = 34 ………(2) (1) – (2), 6d = 24 d=4 By substituting d = 4 into (2), we have 2a + 7(4) = 34 a=3 ∴ The first term is 3 and the common difference is 4. 20 [2(3) + (20 – 1)(4)] 2 = 820

(b) S(20) =

6.

∵ d = –4 and S(8) = 336 8 [2a + (8 – 1)(–4)] = 336 2 a = 56 ∴ The first term is 56.

(b) Let n be the number of terms of the given series. ∵ a = 6, d = 12 – 6 = 6 and l = T(n) = 198 and T(n) = a + (n – 1)d ∴ 198 = 6 + (n – 1)(6) n = 33 33(6 + 198) ∴ The required sum = 2 = 3366 (c) The required sum = sum of integers between 1 and 200 inclusive – sum of integers between 1 and 200 that are multiples of 6 = 20 100 – 3366 (from (a) and (b)) = 16 734 9.

(a) For a common difference of 6, the arithmetic sequence is 2, 8, 14, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(6) = 98 k = 17 ∴ 2 + 8 + 14 + … + 98 = S(17) 17(2 + 98) = 2 = 850 ∴ The sum of the corresponding series is 850. For a common difference of 16, the arithmetic sequence is 2, 18, 34, …, 98. Let 98 be the jth term. i.e. T(j) = 98 2 + (j – 1)(16) = 98 j=7 ∴ 2 + 18 + 34 + … + 98 = S(7) 7(2 + 98) = 2 = 350 ∴ The sum of the corresponding series is 350. (or any other reasonable answers) (b) For a common difference of 8, the arithmetic sequence is 2, 10, 18, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(8) = 98 k = 13

14

∴ 2 + 10 + 18 + … + 98 = S(13) 13( 2 + 98) = 2 = 650 ∴ The sum of the corresponding series is 650. For a common difference of 4, the arithmetic sequence is 2, 6, 10, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(4) = 98 k = 25 ∴ 2 + 6 + 10 + … + 98 = S(25) 25(2 + 98) = 2 = 1250 ∴ The sum of the corresponding series is 1250. (or any other reasonable answers) 10. (a) (–2) + (–1) + 0 + 1 + 2 or (–6) + (–3) + 0 + 3 + 6 (or any other reasonable answers) (b) 1 + 2 + 3 + 4 + 5 or (–1) + 1 + 3 + 5 + 7 (or any other reasonable answers)

Level 2 11. (a) Let k be the number of rows. ∵ a = 12, d = 3 and T(k) = 6a ∴ a + (k – 1)d = 6a 12 + (k – 1)(3) = 6(12) k = 21 ∴ The number of rows is 21. n [2a + (n – 1)d] 2 21 S(21) = [2(12) + (21 – 1)(3)] 2 = 882 ∴ The number of seats is 882.

(b) S(n) =

12. Let m1, m2, m3, …, m25 be the arithmetic means. ∵ a = 5, l = 120 and n = 27 27(5 + 120) ∴ S(27) = 2 i.e. 5 + m1 + m2 + m3 + … + m25 + 120 = 1687.5 m1 + m2 + m3 + … + m25 = 1562.5 ∴ The sum of the 25 arithmetic means between 5 and 120 is 1562.5. 13. Let m1, m2, m3, …, mx be the arithmetic means. ∵ a = x, l = 2x and n = x + 2 ( x + 2)( x + 2 x) ∴ S(x + 2) = 2 ( x + 2)( x + 2 x) i.e. x + m1 + m2 + m3 + … + mx + 2x = 2

80

Arithmetic and Geometric Sequences and their Summation

3x 2 2 ∴ The sum of the x arithmetic means between x and 2x 3x 2 is . 2 m1 + m2 + m3 + … + mx =

14. (a) first term = T(1) = 55 – 2(1) = 53 Common difference = T(n + 1) – T(n) = [55 – 2(n + 1)] – (55 – 2n) = −2 (b) Let the kth term be the first negative term. i.e. T(k) < 0 ∴ 55 – 2k < 0 55 k> 2 ∴ The first negative term is the 28th term. ∴ T(28) = 55 – 2(28) = −1 (c) ∵ T(n) is negative for n ≥ 28. ∴ S(n) is maximum at n = 27. 27 S(27) = [2(53) + (27 – 1)( –2)] 2 = 729 ∴ The maximum value of S(n) is 729. 15. ∵ a = –49, d = (–46) – (–49) = 3 ∴ T(n) = –49 + (n – 1)(3) = 3n – 52 Let the kth term be the last negative term. ∴ T(k) < 0 i.e. 3k – 52 < 0 52 k< 3 ∴ The last negative term is the 17th term. 17 S(17) = [2(–49) + (17 – 1)(3)] 2 = –425 ∴ The sum of all negative terms is –425. 16. (a) The sum = 200 + 205 + 210 + … + 500 ∴ a = 200, d = 5 and l = 500 T(n) = 200 + (n – 1)(5) = 195 + 5n Let k be the number of terms. ∴ T(k) = 500 i.e. 195 + 5k = 500 k = 61 61( 200 + 500) S(61) = 2 = 21 350 ∴ The required sum is 21 350. (b) The sum = 203 + 210 + 217 + … + 497 ∴ a = 203, d = 7 and l = 497 T(n) = 203 + (n – 1)(7) = 196 + 7n Let k be the number of terms.

100

Certificate Mathematics in Action Full Solutions 5A

T(k) = 497 196 + 7k = 497 k = 43 43(203 + 497) S(43) = 2 = 15 050 ∴ The required sum is 15 050. ∴ i.e.

n (1 + n ) 

n (1 + n )  1 +  2   = 2 n(1 + n)  n  = 1 + (1 + n)  4  2  2

(d) The last term in the first (n – 1) brackets = The total number of terms in the first (n – 1) brackets ( n − 1)[1 + ( n − 1)] = 2 n( n − 1) = 2 n(n − 1) Now, a = 1 and l = 2 ∴ The sum of the terms in the first (n – 1) brackets ( n − 1)[1 + ( n − 1)]  (n − 1)  = [1 + ( n − 1)]  1 + 4 2  

(c) If the integers are divisible by both 5 and 7, then they are divisible by 35. We are going to find the sum of integers divisible by 35 between 200 and 500 inclusive. ∴ The sum = 210 + 245 + 280 + … + 490 ∴ a = 210, d = 35 and l = 490 T(n) = 210 + (n – 1)(35) = 175 + 35n Let k be the number of terms. ∴ T(k) = 490 i.e. 175 + 35k = 490 k=9 9(210 + 490) S(9) = 2 = 3150 ∴ The required sum is 3150.

n(n − 1)  n  1 + ( n − 1)  4  2  ∴ The sum of the terms in the nth bracket = The sum of the terms in the first n brackets – the sum of the terms in the first (n – 1) brackets n(1 + n)  n  n(n − 1)  n  = 1 + (1 + n)  − 1 + ( n − 1)  4  2 4  2   n 2 = (1 + n ) 2 =

(d) The required sum = sum of integers divisible by 5 + sum of integers divisible by 7 – sum of integers divisible by both 5 and 7 = 21 350 + 15 050 – 3150 = 33 250 17. 2 × 22 × 23 × …× 264 = 21 + 2 + 3 + …. + 64 64

=22 =2

(1+ 64 )

2080

18. (a) In the 1st bracket, there is 1 term. In the 2nd bracket, there are 2 terms. In the 3rd bracket, there are 3 terms. ∴ Number of terms in the nth bracket is n. (b) The total number of terms = 1 + 2 + 3 + … + n n(1 + n) = 2 (c) The last term in the first n brackets = the total number of terms in the first n brackets n(1 + n) = 2 n(1 + n) Now, a = 1 and l = . 2 ∴ The sum of the terms in the first n brackets

Exercise 14G (p. 196) Level 1 1.

4 = 2 and n = 10 2 10 1( 2 − 1) ∴ S(10) = 2 −1 = 1023

(a) ∵ a = 1, R =

9 1 = and n = 7 27 3   1 7  27 1 −      3  

(b) ∵ a = 27, R =

∴ S(7) =

1 3

1093 = 27 (c) ∵ a = 8, R =

101

1−

−16 = –2 and n = 10 8

14

∴ The number of terms is 6. 1 4 (2 6 − 1) ∴ S(6) = 5 2 −1 3 = 264 5

8[1 − (−2)10 ] ∴ S(10) = 1 − (−2) = − 2728 0.6 = –0.6 and n = 7 −1 − 1[1 − (−0.6) 7 ] ∴ S(7) = 1 − ( −0.6) = − 0.642 496

(d) ∵ a = –1, R =

2.

3.

(a) Let n be the number of terms of the given series. 6 ∵ a = 2, R = = 3 and T(n) = aRn – 1 = 4374 2 ∴ 4374 = 2(3n – 1) n=8 ∴ The number of terms is 8. 2(38 − 1) ∴ S(8) = 3 −1 = 6560 (b) Let N be the number of terms of the given series. −8 ∵ a = 2, R = = –4 and T(N) = aRN – 1 = –2048 2 ∴ –2048 = 2(–4)N – 1 N=6 ∴ The number of terms is 6. 2[1 − ( −4) 6 ] ∴ S(6) = 1 − (−4) = − 1638 (c) Let n be the number of terms of the given series. 1 1 ∵ a = , R =  1  = 3 and T(n) = aRn – 1 = 729 3 3   1 ∴ 729 = (3)n – 1 3 n=8 ∴ The number of terms is 8. 1 8 (3 − 1) ∴ S(8) = 3 3 −1 3280 = 3 (d) Let n be the number of terms of the given series. 1 ∵ a=4 ,R= 5 ∴ 134

80

8 4

2 5 1

= 2 and T(n) = aR

5

2 1 = 4 (2) n −1 5 5 n=6

Arithmetic and Geometric Sequences and their Summation

n–1

2 = 134 5

(a) Let N be the number of terms that must be taken. 2 a ( R N − 1) ∵ a = 1, R = = 2 and S(N) = = 511 1 R −1 1(2 N − 1) ∴ 511 = 2 −1 N=9 ∴ 9 terms of geometric series must be taken. (b) Let N be the number of terms that must be taken. 1 3 1   ∵ a= ,R= = 3 and 1 9 9   S(N) =

a ( R N − 1) 4 = 40 R −1 9 1

N (3 − 1) 4 9 40 = 9 3 −1 N=6 ∴ 6 terms of geometric series must be taken.

∴

(c) Let N be the number of terms that must be taken. 24 1 = ∵ a = 48, R = 48 2 and S(N) =

∴

a (1 − R N ) 29 = 95 1− R 32

  1 N  481 −     2  29  95 =  1 32 1− 2

N = 10 ∴ 10 terms of geometric series must be taken. (d) Let N be the number of terms that must be taken. 2 9 2   ∵ a= ,R= =3  2 27  27    and S(N) =

a ( R N − 1) 26 = 242 R −1 27 2

N (3 − 1) 26 27 242 = 27 3 −1 N=8 ∴ 8 terms of geometric series must be taken.

∴

102

Certificate Mathematics in Action Full Solutions 5A

4.

 5000(1.06)(1.0610 − 1)  S(10) =   1.06 − 1   = 69 858 (cor. to the nearest dollar) ∴ Peter will receive $69 858 at the end of the 10th year.

∵ T(3) = 27 2

 2 ∴ a  −  = 27  3 243 a= 4 10  243  2  1 −  −   4   3   ∴ S(10) =  2 1− −   3 11 605 = 324 5.

Level 2

a ( R 3 − 1) = 12 R −1 ∵ T(4) + T(5) + T(6) = –96 ∴ S(6) – S(3) = –96 a ( R 6 − 1) = –84 R −1

(a) Let N be the number of terms that must be taken. 1 1 − ∵ a= , R =  1  = –3 − 3 3  

……(2)

R6 −1 = –7 R3 −1 R6 + 7R3 – 8 = 0 3 (R + 8)(R3 – 1) = 0 R3 = –8 or R3 = 1 R = –2 or R = 1(rejected) By substituting R = –2 into (1), we have a[(−2) 3 − 1] = 12 ( −2) − 1 a=4 ∴ The first term is 4 and the common ratio is –2. Let n be the least number of terms taken. 6 3 a ( R N − 1) ∵ a = 4, R = = and S(N) = > 800 4 2 R −1  3  N  4   − 1  2   ∴ > 800 3 2

−1

N > 11.38 ∴ At least 12 terms of geometric series must be taken.

103

9.

……(1)

(2) ÷ (1),

7.

At the end of the 1st year, he will get $1000(1.04). At the end of the 2nd year, he will get $1000(1.04) + $1000(1.04)2. At the end of the 3rd year, he will get $1000(1.04) + $1000(1.04)2 + $1000(1.04)3. ∴ a = 1000(1.04), R = 1.04 Let n be the number of years needed. S(n) > 15 000 1000(1.04)(1.04 n − 1)    > 15 000 1.04 − 1   n > 11.6 ∴ The minimum number of years needed is 12.

Let the first term be a and the common ratio be R. ∵ S(3) = 12 ∴

6.

8.

At the end of the 1st year, Peter will receive $5000(1 + 6%)1 = $5000(1.06)1. At the end of the 2nd year, he will receive $5000(1.06) + $5000(1.06)2. At the end of the 3rd year, he will receive $5000(1.06) + $5000(1.06)2 + $5000(1.06)3. ∴ a = 5000(1.06), R = 1.06

and S(N) =

a (1 − R N ) 2 = 60 1− R 3 1

[

]

N − 1 − ( −3) 2 3 60 = 3 1 − (−3) N=6 ∴ 6 terms of geometric series must be taken.

∴

(b) Let N be the number of terms that must be taken. −18 1 =− ∵ a = 36, R = 36 2 and S(N) =

a (1 − R N ) 125 = 23 1− R 128

  1 N  36 1 −  −    2   ∴ 23 125 =  128  1 1− −   2 N = 10 ∴ 10 terms of geometric series must be taken. 10. (a) (i) In the 2nd second, the distance travels = 20 × 0.8 m In the 3rd second, the distance travels = 20 × 0.82 m In the 4th second, the distance travels

14

= 20 × 0.83 m ∴ In the nth second, the distance travels is 20 × 0.8n – 1 m. (ii) ∵ a = 20, R = 0.8 20(1 − 0.8 n ) ∴ S(n) = = 100(1 – 0.8n) 1 − 0.8 ∴ The total distance travels in the first n seconds is 100(1 – 0.8n) m. (b) From (a), the distance travels in the first 18 seconds = S(18) = 100(1 – 0.818) m = 98.2 m < 100 m ∴ The train cannot stop at the station successfully in 18 seconds. 11. (a) After the 1st blow, the length driven is 2 m. After the 2nd blow, the length driven is 2(0.9) m. … After the 6th blow, the length driven is 2(0.9)5 m. ∴ a = 2 and R = 0.9 2(1 − 0.9 6 ) ∴ S(6) = = 9.37, cor. to 2 d.p. 1 − 0. 9 ∴ The length of the pile driven is 9.37 m. 2(1 − 0.9 7 ) m 1 − 0. 9 = 10.43 m (cor. to 2 d.p.) > 10 m ∴ The pile would be completely driven into the ground.

(b) S(7) =

12. (a)

BD = ABsin θ ∴ d1 = x sin θ AD = ABcos θ = xcos θ DE = ADsin θ = xcos θ sin θ ∴ d2 = x sin θ cosθ

(b) (i) AE = ADcos θ = x(cos θ)2 d3 = EF = AEsin θ = xsin θ(cos θ)2 AF = AEcos θ = x(cos θ)3 d4 = FG = AFsin θ = xsin θ(cos θ)3 d 2 x sin θ cosθ = = cosθ d1 x sin θ d 3 x sin θ (cosθ ) 2 = = cosθ d2 x sin θ cosθ d 4 x sin θ (cosθ ) 3 = = cosθ d 3 x sin θ (cosθ ) 2 d 2 d3 d 4 = = d1 d 2 d 3 ∴ d1, d2, d3, d4 are in geometric sequence. ∴

(ii) ∵ a = xcos θ, R = cos θ

80

Arithmetic and Geometric Sequences and their Summation

∴ d1 + d2 + d3 + d4 = S(4) x sin θ [1 − cos 4 θ ] 1 − cosθ = x sin θ (1 + cosθ )(1 + cos 2 θ ) =

(c) d1 + d2 + d3 + d4 = 20sin 30°(1 + cos 30°)(1 + cos230°) 35 = (2 + 3 ) 4 13. (a) 99 – 9 = 90 999 – 99 = 900 ≠ 90 ∴ It is not an arithmetic sequence. 99 = 11 9 999 111 = ≠ 11 99 11 ∴ It is not a geometric sequence. (b) 9 = 10 – 1 = 101 – 1 99 = 100 – 1 = 102 – 1 999 = 1000 – 1 = 103 – 1 ∴ T(n) = 10n – 1 (c) T(1) + T(2) + T(3) + … + T(n) = (101 – 1) + (102 – 1) + (103 – 1) + … + (10n – 1) = 101 + 102 + 103 + … + 10n – n 10(10 n − 1) = −n 10 − 1 10 = (10 n − 1) − n 9 14. (a) First term in the 1st bracket = 1= 20 First term in the 2nd bracket = 2 = 20 + 1 First term in the 3rd bracket = 23 = 20 + 1 + 2 First term in the 4th bracket = 26 = 20 + 1 + 2 + 3 ∴ First term in the nth bracket = 20 + 1 + 2 + … + (n – 1 ) =2

n ( n −1) 2

Last term in the nth bracket 1 = × first term in the (n + 1)th bracket 2 ( n+1)( n +1−1) 1 = ×2 2 2 = 2

n2 +n−2 2

(b) The number of terms in the 1st bracket = 1 The number of terms in the 2nd brackets = 2 The number of terms in the 3rd brackets = 3 ∴ The number of terms in the nth brackets = n

104

Certificate Mathematics in Action Full Solutions 5A

and a = 2

n ( n −1) 2

9

,R=2

∴ The sum =

2

∴ S(∞) = 1 −

n ( n −1) 2

(2 n − 1) 2 −1

= (2 n − 1)2

= 27

n ( n −1) 2

2 1 =− −4 2 −4  1 ∴ S(∞) = 1 −  −   2 8 =− 3

(b) ∵ a = –4, R =

(c) The number of terms in the first bracket = 1 The number of terms in the first 2 brackets =1+2 The number of terms in the first 3 brackets =1+2+3 ∴ The number of terms in the first n brackets = 1 + 2 + 3 +… + n n2 + n = 2 and a = 1, R = 2 ∴ The sum = 1(2 =2

0.2 = 0.2 1 1 ∴ S(∞) = 1 − 0.2 5 = 4

(c) ∵ a = 1, R =

n 2 +n 2

− 1) 2 −1

n 2 +n 2

−1

− 5    1 (d) ∵ a = 5, R =  3  =− 5 3 5 S (∞ ) =  1 ∴ 1− −   3 15 = 4

a (r n − 1) r −1 a 2 [(r 2 ) n − 1] = r 2 −1 S =a +ar +ar +…+ar 2 2 2 2 4 2 2(n – 1) 2 2 a (r 2 n − 1) = r 2 −1

15. S1 = a + ar + ar2 + … + arn – 1 =

(r – 1)S12 + 2aS1 2

 a ( r n − 1)  a( r n − 1)  + 2a = ( r − 1) r −1  r −1  2 n 2 2 n a (r − 1) 2a ( r − 1) = + r −1 r −1 a 2 (r n − 1) n = ( r + 1) r −1 2 2n a (r − 1) = r −1 a 2 (r 2 n − 1) (r + 1) = ⋅ r −1 (r + 1) a 2 (r 2 n − 1) r2 −1 = (r + 1)S2 = ( r + 1)

Exercise 14H (p. 203) Level 1 1.

105

(a) ∵ a = 9 and R =

6 2 = 9 3

2 3

2.

(a) 0.7 = 0.7777… = 0.7 + 0.07 + 0.007 + 0.0007 + … 0.7 = 1 − 0.1 7 = 9 (b) 0.4 7 = 0.474 747… = 0.47 + 0.0047 + 0.000 047 + … 0.47 = 1 − 0.01 47 = 99 (c) 0.23 4 = 0.234 343… = 0.2 + 0.034 + 0.000 34 + 0.000 0034 + … 0.034 = 0.2 + 1 − 0.01 116 = 495

14

(d) 0.7 47 = 0.747 747… = 0.747 + 0.000 747 + 0.000 000 747 + … 0.747 = 1 − 0.001 83 = 111 3.

(a) ∵ ∴

∴

1 a  = 6 2 a = 12 ∴ The first term is 12. 5.

S (∞ ) = 5 2 =5 1− R 3 R= 5

∵ T(1) + T(2) + T(3) = 21 a (1 − R 3 ) = 21 ……(1) 1− R and

∴ The first 3 terms are 2,

(b) ∵

Arithmetic and Geometric Sequences and their Summation

6 18 and . 5 25

S (∞ ) = 24 a = 24 1− R

(1) ÷ (2), 1 – R3 =

40 7 10 40 = 1− R 7 3 R=− 4

S (∞ ) =

By substituting R =

1−

S (∞ ) = 90 a ∴ = 90 1− 0.2 a = 72 ∴ The first 3 terms are 72, 14.4 and 2.88. S (∞ ) = −5 a ∴ = –5 1 − (−0.2) a = –6 ∴ The first 3 terms are –6, 1.2 and –0.24.

6.

2

= 24

2

(a) ∵ a = 1, R = x a 1 ∴ S (∞ ) = 1 − R = 1 − x

∴

7.

……(2) 1 4

1

1 8 1 1 =1 1− x 8 1 x= 9

(b) ∵ S (∞ ) = 1

……(1)

1  R−  = 0 2  1 R= 2 1 By substituting R = into (1), we have 2

80

1 into (2), we have 2

a = 12 ∴ The first 3 terms are 12, 6 and 3.

(d) ∵

(1) ÷ (2), R(1 – R) =

1 2

a

(c) ∵

∵ T(2) = 6 aR = 6 and S (∞ ) = 24 a = 24 1− R

7 8 R=

15 45 ∴ The first 3 terms are 10, − and . 2 8

4.

……(2)

The total possible output of gold = [1000 + 1000(80%) + 1000(80%)(80%) + ….] kg = [1000 + 1000(0.8) + 1000(0.8)(0.8) + ….] kg 1000 = kg 1 − 0.8 = 5000 kg

Level 2 8.

80  2 =  3 120 ∴ Total angle swings through = S (∞ )

(a) a = 120°, R =

106

Certificate Mathematics in Action Full Solutions 5A

=

120  1−

=

2 3

= 360  (b) ∵ Total angle swings through is 360°. ∴ Total distance swings through  360    = (2π)(10 cm)     360  = 20π cm 9.

For the downwards distance travelled, a = 10, R = 75% = 0.75 a m ∴ S (∞ ) = 1− R 10 = m 1 − 0.75 = 40 m For the upwards distance travelled, a = 10(75%) = 7.5, R = 0.75 a 7.5 m= m = 30 m ∴ S (∞ ) = 1− R 1 − 0.75 ∴ The total distance travelled = (30 + 40) m = 70 m

1 3

11. (a) ∵ The speed of Ken is twice that of Angel. ∴ The distance travelled by Ken is twice that of Angel in the same time. 1 1 ∴ BC = AB = (24 m) = 12 m 2 2 1 1 Similarly, CD = BC = (12 m) = 6 m 2 2 1 1 and DE = CD = (6 m) = 3 m 2 2 BC 6 1 = = (b) AB 12 2 CD 3 1 = = BC 6 2 ∴ AB, BC, CD, DE are in geometric sequence with 1 common ratio . 2 1 2 ∴ Total distance Ken must run = S (∞ )

(c) a = 24 m, R =

=

10. (a) (i) The fraction of the original piece of cake P 1 gets the first time = 4

24 1−

1

m

2

= 48 m

(ii) The fraction of the original piece of cake P 11 1 gets the second time alone =   =  4  4 16 (iii) The fraction of the original piece of cake P gets the nth time alone 1 =  4 1 = 2n 2

n −1

1 4

1  16  1 1  = (b) ∵ a = , R = 1 4 4 4   1    4 ∴ S (∞) = 1 1− 4

107

12. (a) ∵ △ABC is an equilateral triangle. ∴ ∠ABC = 60° ∵ Area of △ABC = 3 × Area of △OAB 1 1 ∴ (AB)(BC)(sin ∠ABC) = 3 × (AB)(OD) 2 2 1 1 (8)(8)(sin 60°) = 3 × (8)(r1) 2 2

14

r1=

4 3

Consider △OEF. ∠FOE = 60° OE = OF cos ∠FOE r1 – r2 = (r1 + r2)cos 60° 1 r2 = r1 3 4 r2 = 3 3 Similarly, r3 =

4 1 r2 = 9 3 3

(b) From (a), we know that r1, r2, r3, … are in geometric 4 1 sequence with first term and common ratio . 3 3 Sum of the circumferences = (2πr1 + 2πr2 + 2πr3 + …) cm = 2π(r1 + r2 + r3 + …) cm  4    3 = 2π    cm 1  1 −  3   = 4 3π cm (c) Consider the sequence r12, r22, r32, … r2 2 r12

r =  2  r1

2

 1  = 9 

∴ r12, r22, r32, … are in geometric sequence with 16 1 first term and common ratio . 9 3 Sum of areas of these circles = (πr21 + πr22 + πr32 + …) cm2 = π(r12 + r22 + r32 + …) cm2   16     3 = π    cm 2 1  1− 9    = 6π cm 2 13. (a) C1C = B1C1 = b AC1 = AC – C1C = 3a – b By similar triangles, we have

80

Arithmetic and Geometric Sequences and their Summation

AC1 B1C1 = AC BC 3a − b b = 3a a 3 b= a 4 (b) (i) From (a), we have B1C1 =

3 BC 4

Similarly, we have 3 B2C2 = B1C1 4 3 = b 4

3 b 4 33  = 4  4 a   9 = a 16

(ii) B2C2 =

(c) (i) By considering triangles AB1C1, AB2C2, …and using argument similar to (a) and (b), we have B2 C 2 B3C3 3 = = ... = B1C1 B2 C 2 4 ∴ B1C1, B2C2, B3C3, … are in geometric sequence. 4

3 (ii) B4C4 =   a 4 81 = a 256 (iii) The areas of the squares are in geometric 9 2 sequence with first term a and common 16 9 ratio . 16 Sum of areas = (B1C1)2 + (B2C2)2 + (B3C3)2 +…  9 2  a   =  16 9   1 −    16  9 = a2 7

108

Certificate Mathematics in Action Full Solutions 5A

= (2401 + 1225 + …) cm2    2401   cm 2 = 25   1 −   49   117 649 = cm 2 24

14. (a) A1B1 = B1C1 = 49 cm 3 A2B1 = A1B1 = 21 cm 7 4 B1B2 = B1C1 = 28 cm 7 = A2 B12 + B1 B2 2 cm

AB 2

2

= 212 + 28 2 cm = 35 cm

(Pyth. theorem)

Revision Exercise 14 (p. 209)

A2 B2 5 = (b) Common ratio = A1 B1 7

Level 1 1.

T(3) = 2(3) – 1 = 5 , T(4) = 2(4) – 1 = 7 ,

(c) (i) ∵ The sides of the squares are in geometric sequence. ∴ The perimeters of the squares are also in geometric sequence with the same common ratio and first term 196 cm. Sum of the perimeters = (196 + 140 + … ) cm    196   cm = 1− 5    7   = 686 cm (ii) Notice that 4 A1A2 = A1B1 = 28 cm 7 4 A2A3 = A2B2 = 20 cm 7 … A1A2, A2A3, …are also in geometric sequence with the same common ratio as A1B1, A2B2, … and first term 28 cm. ∴ Total distance travelled by the ant = (28 + 20 + … ) cm    28  cm = 5   1−  7  = 98 cm 2

(d) Common ratio =

A2 B2 2 A1 B1

352 49 2 25 = 49 First term = A1B12 = 492 cm2 = 2401 cm2 ∴ The sum of areas =

109

(a) T(1) = 2(1) – 1 = 1 , T(2) = 2(2) – 1 = 3 , T(5) = 2(5) – 1 = 9 (b) T(1) = 31 – 1 = 1 , T(2) = 32 – 1 = 3 , T(3) = 33 – 1 = 9 , T(4) = 34 – 1 = 27 , T(5) = 35 – 1 = 81 (c) T(1) = (–1)1 + 1 ⋅ 2(1) = 2 , T(2) = (–1)2 + 1 ⋅ 2(2) = −4 , T(3) = (–1)3 + 1 ⋅ 2(3) = 6 , T(4) = (–1)4 + 1 ⋅ 2(4) = −8 T(5) = (–1)5 + 1 ⋅ 2(5) = 10 (d) T(1) = (1)1 + 1 = 1 , T(2) = (2)2 + 1 = 8 , T(3) = (3)3 + 1 = 81 , T(4) = (4)4 + 1 = 1024 , T(5) = (5)5 + 1 = 15 625

2.

(a) (i) 6, 7 (ii) ∵ T(1) = 2 = 1 + 1 T(2) = 3 = 2 + 1 T(3) = 4 = 3 + 1 T(4) = 5 = 4 + 1 ∴ T(n) = n + 1 (b) (i) 20, 24 (ii) ∵ T(1) = 4 = 4(1) T(2) = 8 = 4(2) T(3) = 12 = 4(3) T(4) = 16 = 4(4) ∴ T(n) = 4n (c) (i) 4, 2 (ii) ∵ T(1) = 64 = 27 – 1 T(2) = 32 = 27 – 2

14

T(3) = 16 = 27 – 3 T(4) = 8 = 27 – 4 7−n ∴ T(n) = 2 (d) (i)

(ii) T(9) = 4(9) – 2 = 34 (b) (i) T(n) = –5 + (n – 1)(–4) = −4n − 1

3 3 , 32 64

(ii) T(9) = –4(9) – 1 = −37

3 3 = 2 21 3 3 T(2) = = 2 4 2 3 3 T(3) = = 3 8 2 3 3 = 4 T(4) = 16 2 3 ∴ T(n) = 2 n

(ii) ∵ T(1) =

3.

 1 (c) (i) T(n) = –2 + (n – 1) 1   2 3n − 7 = 2 3(9) − 7 2 = 10

(ii) T(9) =

(a) T(2) – T(1) = 4 – 1 = 3 T(3) – T(2) = 7 – 4 = 3 T(4) – T(3) = 10 – 7 = 3 ∴ It is an arithmetic sequence with common difference 3. T(n) = 1 + (n – 1)(3) = 3n − 2

3  7 (d) (i) T(n) = 4 + (n − 1) − 4    10 − 7 n = 4 10 − 7(9) 4 53 =− 4

(ii) T(9) =

(b) T(2) – T(1) = 7 – 9 = –2 T(3) – T(2) = 5 – 7 = –2 T(4) – T(3) = 2 – 5 = –3 ≠ –2 ∴ It is not an arithmetic sequence. (c)

T(2) – T(1) = log 25 – log 5 = 2 log 5 – log 5 = log 5 T(3) – T(2) = log 125 – log 25 = 3 log 5 – 2 log 5 = log 5 T(4) – T(3) = log 625 – log 125 = 4 log 5 – 3 log 5 = log 5 ∴ It is an arithmetic sequence with common difference log 5. T(n) = log 5 + (n – 1)(log 5) = n log 5

(d) T(2) – T(1) = (x + 3) – (x + 1) = 2 T(3) – T(2) = (x + 5) – (x + 3) = 2 T(4) – T(3) = (x + 7) – (x + 5) = 2 ∴ It is an arithmetic sequence with common difference 2. T(n) = (x + 1) + (n – 1)(2) = x + 2n − 1 4.

(a) (i) T(n) = 2 + (n – 1)(4) = 4n − 2

80

Arithmetic and Geometric Sequences and their Summation

5.

(a) Let a and d be the first term and the common difference respectively. ∵ a = 101 and d = 99 – 101 = –2 ∴ T(n) = 101 + (n – 1)(–2) = 103 – 2n Let –1 be the kth term. i.e. T(k) = –1 ∴ 103 – 2k = –1 k = 52 ∴ There are 52 terms in the sequence. (b) Let a and d be the first term and the common difference respectively. ∵ a = 3 and d = 1 – 3 = –2 ∴ T(n) = 3 + (n – 1)(–2) = 5 – 2n Let –15 be the kth term. i.e. T(k) = –15 ∴ 5 – 2k = –15 k = 10 ∴ There are 10 terms in the sequence. (c) Let a and d be the first term and the common difference respectively. 1 1 1 ∵ a = and d = 1 − = 2 2 2

110

Certificate Mathematics in Action Full Solutions 5A

∴ T(n) = a + (n – 1)d 125  5 = + (n − 1) −  2  2 5 = 65 − n 2

1 1 + (n – 1)   2 2 1 = n 2 Let 10 be the kth term. i.e. T(k) = 10 1 ∴ k = 10 2 k = 20 ∴ There are 20 terms in the sequence. ∴ T(n) =

(c) Let a and d be the first term and the common difference respectively. T(4) = a + 3d = 75 ……(1) T(10) = a + 9d = 117 ……(2) (2) – (1), 6d = 42 d=7 By substituting d = 7 into (1), we have a + 3(7) = 75 a = 54 ∴ T(n) = a + (n – 1)d = 54 + (n – 1)(7) = 47 + 7 n

(d) Let a and d be the first term and the common difference respectively. 1 2 1 ∵ a = and d = − − = –1 3 3 3 1 ∴ T(n) = + (n – 1)(–1) 3 4 = –n 3 2 Let − 18 be the kth term. 3 2 i.e. T(k) = − 18 3 4 2 − k = −18 ∴ 3 3 k = 20 ∴ There are 20 terms in the sequence. 6.

(a) Let a and d be the first term and the common difference respectively. T(7) = a + 6d = 20 ……(1) T(19) = a + 18d = 56 ……(2) (2) – (1), 12d = 36 d=3 By substituting d = 3 into (1), we have a + 6(3) = 20 a=2 ∴ T(n) = a + (n – 1)d = 2 + (n – 1)(3) = 3n − 1 (b) Let a and d be the first term and the common difference respectively. T(10) = a + 9d = 40 ……(1) T(16) = a + 15d = 25 ……(2) (2) – (1), 6d = –15 5 d=− 2 5 By substituting d = − into (1), we have 2  5 a + 9 −  = 40  2 125 a= 2

(d) Let a and d be the first term and the common difference respectively. T(7) = a + 6d = 62 ……(1) T(19) = a + 18d = 2 ……(2) (2) – (1), 12d = –60 d = –5 By substituting d = –5 into (1), we have a + 6(–5) = 62 a = 92 ∴ T(n) = a + (n – 1)d = 92 + (n – 1)(–5) = 97 − 5n 7.

− 8 + ( −2) 2 = –5

(a) Arithmetic mean =

117 + 49 2 = 83

(b) Arithmetic mean =

− 13 + 13 2 =0

(c) Arithmetic mean =

8.

(a) Let d1 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 5, 5 + d1, 5 + 2d1, 17 ∵ The 4th term is also given by 5 + 3d1. ∴ 5 + 3d1 = 17 d1 = 4 ∴ The two required arithmetic means are 9 and 13. (b) Let d2 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is:

14

3, 3 + d2, 3 + 2d2, 3 + 3d2, 19 ∵ The 5th term is also given by 3 + 4d2. ∴ 3 + 4d2 = 19 d2 = 4 ∴ The three required arithmetic means are 7, 11 and 15. (c) Let d be the common difference of the arithmetic sequence. ∵ T(1) = a and T(4) = b ∴ b = a + 3d b−a d= 3 (i) x = T(2) =a+d b−a = a + 3    2a + b = 3 (ii) y = T(3) = a + 2d b−a = a + 2 3    a + 2b = 3 9.

(a)

T ( 2) −2 1 = = T (1) − 4 2 T (3) −1 1 = = T ( 2) − 2 2

(b)

T (2) 5 = T (1) 2 T (3) 11 5 = ≠ T ( 2) 5 2

T (2) 0.33 11 = = T (1) 0.3 10 T (3) 0.333 111 11 = = ≠ T ( 2) 0.33 110 10

80

(d)

T (2) x 3 y 2 x = = T (1) x 2 y 3 y T (3) x4 y x = 3 2 = T (2) x y y T (4) x5 x = 4 = T (3) x y y ∴ It is a geometric sequence with common x ratio . y n−1

2 3 x  ∴ T(n) = x y    y n+1 = x ⋅ y 4− n

10. (a) (i) ∵ a = 1 and R = 2 and T(n) = aRn – 1 ∴ T(n) = 1(2)n – 1 n −1 =2 (ii) T(10) = 210−1 9 = 512 (or 2 )

(ii) T(10) = 2(–3)10 – 1 −39 366 (or − 2 ⋅ 39 ) = (c) (i) ∵ a = 1 and R = –1 and T(n) = aRn – 1 ∴ T(n) = 1(–1) n – 1 n−1 = ( −1)

n −1

∴ It is not a geometric sequence. (c)

∴ It is not a geometric sequence.

(b) (i) ∵ a = 2 and R = –3 and T(n) = aRn – 1 n −1 ∴ T(n) = 2 ⋅ ( −3)

− 1  T ( 4)  2  1 = = T (3) −1 2 ∴ It is a geometric sequence with common 1 ratio . 2 1 ∴ T(n) = − 4  2 = − 2 3− n

Arithmetic and Geometric Sequences and their Summation

(ii) T(10) = (–1)10 – 1 = −1 1 and R = –2 and T(n) = aRn – 1 4 1 n −1 ∴ T(n) = − 4 (−2) = (−1) n ⋅ 2 n−3

(d) (i) ∵ a = −

(ii) T(10) = (−1)10 (210−3 ) 7 = 128 (or 2 ) 11. (a) Let a and R be the first term and the common ratio respectively.

112

Certificate Mathematics in Action Full Solutions 5A

∵ a = 1 and R =

2 =2 1

∴ T(n) = aRn – 1 = 1(2)n – 1 = 2n – 1 Let 2048 be the kth term. i.e. T(k) = 2048 2k – 1 = 2048 k = 12 ∴ The number of terms is 12. (b) Let a and R be the first term and the common ratio respectively. −10 ∵ a = –2 and R = =5 −2 ∴ T(n) = aRn – 1 = –2(5)n – 1 Let –6250 be the kth term. i.e. T(k) = –6250 –2(5)k – 1 = –6250 k=6 ∴ The number of terms is 6.

(2) ÷ (1),

R5 = 243 R=3 By substituting R = 3 into (1), we have 1 a(3) = 9 1 a= 27 ∴ T(n) = aRn – 1 1 = ⋅ 3n – 1 27 =3

n−4

(b) Let a and R be the first term and the common ratio respectively. T(4) = aR3 = 2 ………(1) 1 T(5) = aR4 = ………(2) 2 1 (2) ÷ (1), R= 4 1 By substituting R = into (1), we have 4 3

(c) Let a and R be the first term and the common ratio respectively. 1.6 1 = ∵ a = 3.2 and R = 3.2 2 n–1 ∴ T(n) = aR n −1

1 = 3.2   2 Let 0.006 25 be the kth term. i.e. T(k) = 0.006 25 1 3.2   2

k −1

= 0.006 25

k = 10 ∴ The number of terms is 10. (d) Let a and R be the first term and the common ratio respectively. −12 ∵ a = 3 and R = = –4 3 n–1 ∴ T(n) = aR = 3(–4)n – 1 Let 12 288 be the kth term. i.e. T(k) = 12 288 3(–4)k – 1 = 12 288 k=7 ∴ The number of terms is 7. 12. (a) Let a and R be the first term and the common ratio respectively. 1 T(2) = aR = ………(1) 9 T(7) = aR6 = 27 ………(2)

113

1 a  = 2 4 a = 128 ∴ T(n) = aRn – 1 1 = 128   4 =2

n −1

9−2 n

(c) Let a and R be the first term and the common ratio respectively. T(3) = aR2 = –8 ………(1) T(6) = aR5 = 1 ………(2) 1 (2) ÷ (1), R3 = − 8 1 R=− 2 1 By substituting R = − into (1), we have 2 2

 1 a  −  = –8  2 a = –32 ∴ T(n) = aRn – 1  1 = –32  −   2 n ( − 1 ) ⋅ 26− n =

n −1

(d) Let a and R be the first term and the common ratio respectively. T(2) = aR = 6 ………(1) T(7) = aR6 = 192 ………(2)

14

(2) ÷ (1),

(d) ∵ first term = (a – b), d = (3a + b) – (a – b) = 2a + 2b and n = 10 10 ∴ S(10) = [2(a – b) + (10 – 1)(2a + 2b)] 2 = 100a + 80b

R5 = 32 R=2 By substituting R = 2 into (1), we have a(2) = 6 a=3 ∴ T(n) = aRn – 1 n−1 = 3⋅ 2

15. (a) (i) Let n be the number of terms of the given series. ∵ a = 200, l = 2 and S(n) = 10 100 n( 200 + 2) ∴ = 10 100 2 n = 100 ∴ There are 100 terms in the given series.

13. (a) Geometric mean = 4 × 64 = 16 (b) Geometric mean = ± − 3 × (−27) = 9 or − 9 (c) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 1 1 1 2 27 , R, R , 4 4 4 32 1 ∵ The 4th term is also given by R3. 4 1 3 27 R = ∴ 4 32 3 R= 2 ∴ The two required geometric means are

(ii) Let d be the common difference. ∵ l = T(100) = 2 ∴ 200 + 99d = 2 d = –2 ∴ The common difference is –2.

3 and 8

9 . 16 (d) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 3, 3r, 3r2, 24 ∵ The 4th term is also given by 3r3. ∴ 3r3 = 24 r=2 ∴ a = 6 and b = 12 14. (a) ∵ a = –3, d = 2 – (–3) = 5 and n = 20 20 ∴ S(20) = [2(–3) + (20 – 1)(5)] 2 = 890 (b) ∵ a = 65, d = 62 – 65 = –3 and n = 15 15 ∴ S(15) = [2(65) + (15 – 1)(–3)] 2 = 660 (c) ∵ a = 7, d = 4 – 7 = –3 and n = 12 12 ∴ S(12) = [2(7) + (12 – 1)( –3)] 2 = −114

80

Arithmetic and Geometric Sequences and their Summation

(b) (i) Let a be the first term and d be the common difference. ∵ S(9) = 90 9 ∴ (2a + 8d) = 90 2 a + 4d = 10 ………(1) ∵ T(11) = 14 a + 10d = 14 ………(2) (2) – (1), 6d = 4 2 d= 3 2 ∴ The common difference is . 3 (ii) By substituting d = a=

2 into (1), we have 3

22 3

50   22   2  ∴ S(50) = 2 2 3  + 49 3       3550 = 3 16. (a) ∵ a = 1, l = 50 50 ∴ S(50) = [1 + 50] 2 = 1275 (b) The sum of even integers = 2 + 4 + 6 +…. + 100 = 2( 1 + 2 + 3 + … + 50) = 2(1275) (from (a)) = 2550

114

Certificate Mathematics in Action Full Solutions 5A

6 17. (a) ∵ a = 3, R = = 2 and n = 8 3 3( 28 − 1) ∴ S(8) = 2 −1 = 765 5 1 = and n = 6 10 2   1 6  101 −      2   ∴ S(6) = 1 1− 2 11 = 19 16

4n – 1 > 16425 n > 7.002 ∴ 8 terms of geometric series must be taken. 19. (a) ∵ a = 28 and R =

∴

(b) ∵ a = 10, R =

(c) ∵ a = 8, R =

−24 = –3 and n = 8 8

8[1 − (−3)8 ] ∴ S(8) = 1 − ( −3) = − 13 120 6 1 = − and n = 12 − 18 3   1 12  − 181 −  −     3   ∴ S(12) =  1 1− −   3 265 720 =− 19 683

(d) ∵ a = –18, R =

18. (a) Let N be the number of terms of the given series. 3 ∵ a = 1, R = = 1 1 and S(N) > 2000 1(3 N − 1) ∴ > 2000 3 −1 3N – 1 > 4000 N > 7.55 ∴ 8 terms of geometric series must be taken. (b) Let n be the number of terms that must be taken. 8 3 2   ∵ a= ,R= =4 2 3 3   and S(N) > 3650 2

∴

115

3

(4 n − 1) 4 −1

> 3650

14 1 = 28 2

28 1 1− 2 = 56

S (∞ ) =

−18 2 =− 27 3 27 S (∞ ) =  2 ∴ 1− −   3 81 = 5

(b) ∵ a = 27, R =

− 9   25  3   =−3 (c) ∵ a = , R = 5 3 5 5    3   5 S (∞ ) =   ∴  3 1− −   5 3 = 8 20. (a) 0.5 = 0.5555… = 0.5 + 0.05 + 0.005 + 0.0005 + … 0.5 = 1 − 0.1 5 = 9 (b) 0.41 4 = 0.414 1414… = 0.4 + 0.014 + 0.000 14 + … 0.014 = 0.4 + 1 − 0.01 41 = 99 (c) 0.4 14 = 0.414 414… = 0.414 + 0.000 414 + 0.000 000 414 + … 0.414 = 1 − 0.001 46 = 111

Level 2

14

21. ∵ The equation ax2 +2bx + c = 0 has equal roots. ∴ ∆=0 i.e. (2b)2 – 4ac = 0 b2 = ac c b = b a ∴ a, b, c are in geometric sequence. 22.

θ = 30

or 

sinθ =

1 2

 or 150

23. (a) a ⋅ a2 ⋅ a3 ⋅ a3 ⋅ …⋅ a100 = a1 + 2 + 3 + … + 100 =a

100 (1+100 ) 2

= a 5050 (b) (2 + log 2) + (4 + log 4) + (8 + log 8) + … = (2 + 4 + 8 + …) + (log 2 + log 4 + log 8 + …) Then we have a geometric sequence 2, 4, 8, … and an arithmetic sequence log 2, log 4, log 8, … For the geometric sequence 2, 4, 8, …, a = 2, R = 2, 2(210 − 1) S(10) = 2 −1 = 2046 For the arithmetic sequence log 2, log 4, log 8, … a = log 2, d = log 2, 10 S(10) = (log 2 + 10 log 2) 2 = 55 log 2 ∴ The required sum = 2046 + 55 log 2 10a 24. (a) (i) Common ratio = a = 10 a(10 n − 1) 10 − 1 a = (10 n − 1) 9

(ii) S(n) =

(b) (i) T(2) – T(1) = log 10a – log a = log 10 = 1 T(3) – T(2) = log 100a – log 10a = log 10 = 1 … ∴ log a, log 10a, log 100a, … are in arithmetic

80

sequence. n [2 log a + (n – 1)(1)] 2 n(n − 1) = n log a + 2

(ii) S(n) =

25. (a) The weight loss in a month is a geometric sequence with a = 2 and R = 80% = 0.8. n −1 ∴ In the nth month, weight loss = 2 ⋅ 0.8 kg

tan θ cosθ = 2 sin θ tan θ cos θ = 2 sin2θ sin θ = 2 sin2θ sin θ (2 sinθ – 1) = 0 sin θ = 0

Arithmetic and Geometric Sequences and their Summation

(b) Total weight loss = S (∞) 2 = kg 1 − 0.8 = 10 kg ∴ Her ultimate weight = (80 – 10) kg = 70 kg 26. ∵ The diameters of the semi-circles are in geometric sequence with a = 8 mm, R = 60% = 0.6 ∴ The lengths of the semi-circles are in geometric 8 sequence with a = π mm = 4π mm, R = 0.6 2 Maximum length = S (∞ ) 4π = mm 1 − 0.6 = 10π mm sin 2 α cos 2 α sin 2 α 2 = cos α

27. (a) Common ratio =

(b) S(∞) =

sin 2 α 1 − cos 2 α

=1 3 1 = 9 3 9 27 = = 1 S(∞) = 2 1−

28. (a) a = 9, R =

3

5  1 9 1 −     3  S(5) =     121 = 1 9 1− 3

∴ Error = S(5) – S (∞) 121 27 = − 2 9 1 =− 18

116

Certificate Mathematics in Action Full Solutions 5A

=

 1 −  18  = × 100% (b) Percentage error  27     2  = –0.412%

=

= 8 2 cm Side of the 3rd square 2

8 2  8 2     =   2  +  2  cm     = 8 cm

2

(ii) The last deposit will amount to 1.03x. The total sum = $[x(1.03)20 + x(1.03)19 +… + 1.03x] 1.03 x(1.0320 − 1) =$ 1.03 − 1 103x (1.0320 − 1) =$ 3

Side of the 1st square = 16 cm Side of the 2nd square = 8 2 cm  1  = 16  cm  2 Side of the 3rd square = 8 cm 2

 1  = 16  cm  2  1   ∴ Side of the kth square = 16   2

k-1

cm

(b) The lengths of each side of the squares are in 1 geometric sequence with a = 16 cm, R = . 2 The perimeter of the squares are also in geometric 1 sequence with a = 4 × 16 cm = 64 cm, R = . 2   1 10  641 −      2   ∴ S(10) = cm 1 1− 2 = 62(2 + 2 ) cm (c) The areas of the squares are in geometric sequence 1 with a = (16 × 16) cm2 = 256 cm2 and R = . 2 Total area of the infinite number of squares formed

117

cm 2

30. (a) (i) From 1st January 2010 to 31st December 2019, there are 20 half-years. The 1st deposit will amount to x(1.03)20. The 2nd deposit will amount to x(1.03)19. The 3rd deposit will amount to x(1.03)18. ∴ The required sum = $[x(1.03)20 + x(1.03)19 + x(1.03)18] 20 19 18 = $ x (1.03 + 1.03 + 1.03 )

2

16 16 =   +   cm 2  2

2

1−

1

= 512 cm 2

29. (a) Side of the 2nd square 2

a 1− R 256

(b) The sum = 500 000 103 x(1.0320 − 1) = 500 000 3 x = 18 066 (cor. to the nearest integer) 31. ∵ a, b, c are in arithmetic sequence. ∴ b–a=c–b b2(c + a) – a2(b + c) = b2c + b2a – a2b – a2c = c(b2 – a2) + ab(b – a) = (b – a)[c(b + a) + ab] = (b – a)(cb + ca + ab) c2(a + b) – b2(c + a) = c2a + c2b – b2c – b2a = a(c2 – b2) + bc(c – b) = (c – b)[a(c + b) + bc] = (c – b)(ac + ab + bc) = (b – a)(cb + ca + ab) = b2(c + a) – a2(b + c) 2 2 ∴ a (b + c), b (c + a), c2(a + b) are in arithmetic sequence. 32. (a) OA2 = OB1cos θ = kcos θ OA3 = OB2cos θ 2 = kcos θ

14

∩

(b) (i)

Arithmetic and Geometric Sequences and their Summation

πθ 180 kπθ = 180

1

A1 B1 = OA1

=

1 − cos 2 θ k cosθ sin θ = 2 sin 2 θ 2 k cosθ = 2 sin θ k2 = 2 tan θ

πθ 180 πkθ cos θ = 180 ∩ πθ A3 B3 = OA3 180 πkθ cos 2 θ = 180 ∩

∩

∩

A1 B1 + A2 B2 + A3 B3 + ...  kπθ    180   = 1 − cosθ kπθ = 180 (1 − cosθ )

(c) Area of △OA2B1 =

Area of △OA3B2 =

Area of △OA4B3 =

1 (OB1)(OA2)sin θ 2 1 = (k)(kcos θ)sin θ 2 1 2 = 2 k cos θ sin θ 1 (OB2)(OA3)sin θ 2 1 = (kcos θ)(kcos2θ)sin θ 2 1 2 3 = 2 k cos θsin θ 1 (OB3)(OA4)sin θ 2 1 = (kcos2θ)(OB3cos θ)sin θ 2 1 = (kcos2θ)(kcos2θcos θ)sin θ 2 1 2 5 = 2 k cos θsin θ

(d) The areas of the triangles are in geometric sequence 1 with a = k2cos θsin θ and R = cos2θ. 2 ∴ The sum to infinity of the series

80

8 8 8 (9) = (10 – 1) = (101 – 1) 9 9 9 8 8 8 88 = (99) = (100 – 1) = (102 – 1) 9 9 9 8 8 8 888 = (999) = (1000 – 1) = (103 – 1) 9 9 9 8 n ∴ T(n) = 9 (10 − 1)

33. (a) 8 =

(ii) The sum is a geometric series with kπθ a= and R = cos θ. 180 ∩

k 2 cosθ sin θ 2

A2 B2 = OA2

∴

2

(b) T(1) + T(2) + T(3) + … + T(n) 8 8 8 = (101 – 1) + (102 – 1) + … + (10n – 1) 9 9 9 8 1 2 n = [10 – 1 + 10 – 1 + … + 10 – 1] 9 8 = [101 + 102 + … + 10n – n] 9 =

 8 10(10 n − 1) − n  9  (10 − 1) 

=

 8 10(10 n − 1) − n  9  9 

34. (a) (i) a4 = 1 + 2(1) + 2(2) + 2(3) = 13 a5 = 1 + 2(1) + 2(2) + 2(3) + 2(4) = 21 (ii) l = 2(n − 1) an = 1 + 2(1) + 2(2) + … + l = 1 + 2(1) + 2(2) + … + 2(n – 1) = 1 + 2[1 + 2 + … + (n – 1)]  n( n − 1)  = 1 + 2   2  = 1 + n( n − 1) (b) (i) 1 = 2(1) – 1 5 = 2(1 + 2) – 1 11 = 2(1 + 2 + 3) – 1 19 = 2(1 + 2 + 3 + 4) – 1 ∴ L = 2(1 + 2 + 3 + … + n) – 1

118

Certificate Mathematics in Action Full Solutions 5A

 n(n + 1)  = 2 –1  2  = n( n + 1) − 1

Multiple Choice Questions (p. 216) 1.

Answer: B Let d be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: x, x + d, x + 2d, x + 3d, x + 4d, x + 5d, y ∵ The 7th term is also given by x + 6d. ∴ x + 6d = y y−x d= 6 x+ y ∴ The 3rd arithmetic mean is . 2

2.

Answer: B d = n2 – n1 Common difference of the new arithmetic sequence = (3n2 + 1) – (3n1 + 1) = 3(n2 – n1) = 3d

3.

Answer: C T(7) = S(7) – S(6) = [3(7)2 – 2(7)] – [3(6)2 – 2(6)] = 133 – 96 = 37

4.

Answer: C

(ii) The terms in bn are in arithmetic sequence with a = an = 1 + n(n – 1) and d = 2. The number of terms = n ∴ The sum of terms in bn = an + (an + 2) + (an + 4) + … + L n = [(1 + n(n – 1)) + (n(n + 1) – 1)] 2 3 =n 35. No Let the speed of the tortoise be v m/s, then that of Achilles be 5v m/s. ∵ The speed of Achilles is 5 times that of the tortoise. ∴ Distance travelled by Achilles is 5 times that of the tortoise. ∴ When Achilles reaches B, the tortoise move 2 m. i.e. BC = 2 m 1  Similarly, CD = BC  m  5  2 m 5 ∴ AB, BC, CD, … are in geometric sequence with a = 1 10 m and R = . 5 ∴ Total distance travelled by Achilles when he reaches the tortoise 10 m 1 = 1− =

10 ⋅ 102 ⋅ 103 ⋅…⋅ 10n > 1050 101 + 2 + 3 + … + n > 1050 1 + 2 + 3 + … + n > 50 n( n + 1) > 50 2 n(n + 1) > 100 ∴ (n + 1)(n + 1) > 100 n + 1 > 10 or ∴ n > 9 or The smallest value of n is 10.

5

= 12.5 m 12.5 5 = s , which is finite. ∴ The time needed = 5v 2v 36. (a) 1, 2, 3 or 2, 4, 6 or 3, 6, 9 (or any other reasonable answers)

5.

(b) By adding the corresponding terms of 1, 2, 3 and 2, 4, 6, we have 3, 6, 9, which is an arithmetic sequence with a common difference 3. The common difference of the new sequence is the sum of that of the former sequences.

(b) By multiplying the corresponding terms of 1, 2, 4 and 1, 3, 9, we have 1, 6, 36, which is a geometric sequence with a common ratio 6. The common ratio of the new sequence is the product of that of the former sequences.

119

Answer: A Let d1 and d2 be the common differences of the arithmetic sequences respectively. Then b = a + 3d1 and b = a + 4d2 m1 + m2 a + d1 + a + 2d1 = n1 + n3 a + d 2 + a + 3d 2 =

37. (a) 1, 2, 4 or 1, 3, 9 or 1, 4, 16 (or any other reasonable answers)

n + 1 < –10 n < –11(rejected)

a + a + 3d1 a + a + 4d 2

a+b a+b =1

=

6.

Answer: C Consider the sequence which terms that are multiples of 7: 7, 14, 21, … Let n be the number of multiples of 7 less than 100.

14

T(n) < 100 7 + (n – 1)(7) < 100 n < 14.29 ∴ The number of multiples of 7 less than 100 is 14. ∴ The number of positive integers less than 100 not divisible by 7 = 99 – 14 = 85 7.

Answer: C Consider the geometric sequence: 1, 2, 4 and 1 + 10, 2 + 10, 4 + 10 = 11, 12, 14 which is not a geometric sequence. ∴ I is false. ∵ a, b , c are in geometric sequence. b c = =R ∴ a b 10b b = =R ∵ 10a a 10c c = =R and 10b b ∴ 10a, 10b , 10c are in geometric sequence. ∴ II is true. ∵ and

b a

=

b = R a

1 10 1 xy = 100

xy =

log x + log y = log xy 1 = log 100 = –2 10. Answer: D 1[(9 2 ) n+1 − 1] 92 − 1 2 2 n+2 −1 1 + 9 + 9 + … +9 = (3 ) 2 4 2n 80 34 n + 4 − 1 = 80 =

11. Answer: B

12. Answer: D S (∞ ) = 100 a = 100 1− 0.25 a = 75

c

c = = R b b

∴ a , b , c are in geometric sequence. ∴ III is true. 8.

Answer: C For I: 1  1   33  1  333  11 1  =  = ≠ and  1  11  1  111 11 3  33      ∴ I is not in geometric sequence.

HKMO (p. 217) 1.

Consider the sequence of the denominator. 2, 3, 3, 4, 4, 4, … , n, n, … ,n, … For each n, 1. there are (n – 1) terms. 2. the terms are from [1 + 2 + 3 +… + (n – 2) + 1]th to [1 + 2 + 3 +… + (n – 1)]th ( n − 1)(n − 2) n( n − 1) + 1 ]th to i.e. from [ th 2 2 If the denominator is 46, the terms are from [ ( 46 − 1)(46 − 2) 46(46 − 1) + 1 ]th to th 2 2 i.e. from 991th to 1035th. For each denominator with value of n, the numerator runs from 1 to n – 1. By counting from the 991th term, the numerator of the 1001th term is 11.

2.

∵ sin 30° + sin230° + … + sinQ30° = 1– cosR45°

For II: 0.33 11 0.333 111 11 = = ≠ and 0.3 10 0.33 110 10 ∴ II is not in geometric sequence. For III: − 1  1 − 1   6    24    = − 1  12  = − 1   =−1 , and 1 1 1 2 −  2 2     3  6  12        ∴ III is in geometric sequence. 9.

Arithmetic and Geometric Sequences and their Summation

Answer: D

80

120

Certificate Mathematics in Action Full Solutions 5A

∴

sin 30  (1 − sin Q 30  ) = 1 − cos R 45 1 − sin 30  0.5(1 − sin Q 30  ) = 1 − cos R 45 0.5 1 – sinQ30° = 1– cosR45° sinQ30° = cosR45° Q  1  1    =    2  2

 1     2   ∴

121

2Q

 1   =    2 R = 2Q

R

R